projectile motion experiment error

5.3 Projectile Motion

Section learning objectives.

By the end of this section, you will be able to do the following:

• Describe the properties of projectile motion
• Apply kinematic equations and vectors to solve problems involving projectile motion

Teacher Support

The learning objectives in this section will help your students master the following standards:

• (C) analyze and describe accelerated motion in two dimensions using equations.

In addition, the High School Physics Laboratory Manual addresses content in this section in the lab titled: Motion in Two Dimensions, as well as the following standards:

• (C) analyze and describe accelerated motion in two dimensions using equations, including projectile and circular examples.

Section Key Terms

Properties of Projectile Motion

Projectile motion is the motion of an object thrown (projected) into the air when, after the initial force that launches the object, air resistance is negligible and the only other force that object experiences is the force of gravity. The object is called a projectile , and its path is called its trajectory . Air resistance is a frictional force that slows its motion and can significantly alter the trajectory of the motion. Due to the difficulty in calculation, only situations in which the deviation from projectile motion is negligible and air resistance can be ignored are considered in introductory physics. That approximation is often quite accurate.

[BL] [OL] Review addition of vectors graphically and analytically.

[BL] [OL] [AL] Explain the term projectile motion. Ask students to guess what the motion of a projectile might depend on? Is the initial velocity important? Is the angle important? How will these things affect its height and the distance it covers? Introduce the concept of air resistance. Review kinematic equations.

The most important concept in projectile motion is that when air resistance is ignored, horizontal and vertical motions are independent , meaning that they don’t influence one another. Figure 5.27 compares a cannonball in free fall (in blue) to a cannonball launched horizontally in projectile motion (in red). You can see that the cannonball in free fall falls at the same rate as the cannonball in projectile motion. Keep in mind that if the cannon launched the ball with any vertical component to the velocity, the vertical displacements would not line up perfectly.

Since vertical and horizontal motions are independent, we can analyze them separately, along perpendicular axes. To do this, we separate projectile motion into the two components of its motion, one along the horizontal axis and the other along the vertical.

We’ll call the horizontal axis the x -axis and the vertical axis the y -axis. For notation, d is the total displacement, and x and y are its components along the horizontal and vertical axes. The magnitudes of these vectors are x and y , as illustrated in Figure 5.28 .

As usual, we use velocity, acceleration, and displacement to describe motion. We must also find the components of these variables along the x - and y -axes. The components of acceleration are then very simple a y = – g = –9.80 m/s 2 . Note that this definition defines the upwards direction as positive. Because gravity is vertical, a x = 0. Both accelerations are constant, so we can use the kinematic equations. For review, the kinematic equations from a previous chapter are summarized in Table 5.1 .

Where x is position, x 0 is initial position, v is velocity, v avg is average velocity, t is time and a is acceleration.

Solve Problems Involving Projectile Motion

The following steps are used to analyze projectile motion:

• Separate the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ A x = A cos θ and A y = A sin θ A y = A sin θ are used. The magnitudes of the displacement s s along x- and y-axes are called x x and y . y . The magnitudes of the components of the velocity v v are v x = v ​ ​ ​ cos θ v x = v ​ ​ ​ cos θ and v y = v ​ ​ ​ sin θ v y = v ​ ​ ​ sin θ , where v v is the magnitude of the velocity and θ θ is its direction. Initial values are denoted with a subscript 0.
• Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms Horizontal Motion ( a x = 0 ) x = x 0 + v x t v x = v 0 x = v x = velocity  is a constant. Horizontal Motion ( a x = 0 ) x = x 0 + v x t v x = v 0 x = v x = velocity  is a constant. Vertical motion (assuming positive is up a y = − g = − 9.80  m/s 2 a y = − g = − 9.80  m/s 2 ) y = y 0 + 1 2 ( v 0 y + v y ) t v y = v 0 y − g t y = y 0 + v 0 y t − 1 2 g t 2 v y 2 = v 0 y 2 − 2 g ( y − y 0 ) y = y 0 + 1 2 ( v 0 y + v y ) t v y = v 0 y − g t y = y 0 + v 0 y t − 1 2 g t 2 v y 2 = v 0 y 2 − 2 g ( y − y 0 )
• Solve for the unknowns in the two separate motions (one horizontal and one vertical). Note that the only common variable between the motions is time t t . The problem solving procedures here are the same as for one-dimensional kinematics.

Teacher Demonstration

Demonstrate the path of a projectile by doing a simple demonstration. Toss a dark beanbag in front of a white board so that students can get a good look at the projectile path. Vary the toss angles, so different paths can be displayed. This demonstration could be extended by using digital photography. Draw a reference grid on the whiteboard, then toss the bag at different angles while taking a video. Replay this in slow motion to observe and compare the altitudes and trajectories.

Tips For Success

For problems of projectile motion, it is important to set up a coordinate system. The first step is to choose an initial position for x x and y y . Usually, it is simplest to set the initial position of the object so that x 0 = 0 x 0 = 0 and y 0 = 0 y 0 = 0 .

Watch Physics

Projectile at an angle.

This video presents an example of finding the displacement (or range) of a projectile launched at an angle. It also reviews basic trigonometry for finding the sine, cosine and tangent of an angle.

• The time to reach the ground would remain the same since the vertical component is unchanged.
• The time to reach the ground would remain the same since the vertical component of the velocity also gets doubled.
• The time to reach the ground would be halved since the horizontal component of the velocity is doubled.
• The time to reach the ground would be doubled since the horizontal component of the velocity is doubled.

Worked Example

A fireworks projectile explodes high and away.

During a fireworks display like the one illustrated in Figure 5.30 , a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75° above the horizontal. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

The motion can be broken into horizontal and vertical motions in which a x = 0 a x = 0 and   a y = g   a y = g . We can then define x 0 x 0 and y 0 y 0 to be zero and solve for the maximum height .

By height we mean the altitude or vertical position y y above the starting point. The highest point in any trajectory, the maximum height, is reached when   v y = 0   v y = 0 ; this is the moment when the vertical velocity switches from positive (upwards) to negative (downwards). Since we know the initial velocity, initial position, and the value of v y when the firework reaches its maximum height, we use the following equation to find y y

Because y 0 y 0 and v y v y are both zero, the equation simplifies to

Solving for y y gives

Now we must find v 0 y v 0 y , the component of the initial velocity in the y -direction. It is given by v 0 y = v 0 sin θ v 0 y = v 0 sin θ , where v 0 y v 0 y is the initial velocity of 70.0 m/s, and θ = 75 ∘ θ = 75 ∘ is the initial angle. Thus,

Since up is positive, the initial velocity and maximum height are positive, but the acceleration due to gravity is negative. The maximum height depends only on the vertical component of the initial velocity. The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding.

There is more than one way to solve for the time to the highest point. In this case, the easiest method is to use y = y 0 + 1 2 ( v 0 y + v y ) t y = y 0 + 1 2 ( v 0 y + v y ) t . Because y 0 y 0 is zero, this equation reduces to

Note that the final vertical velocity, v y v y , at the highest point is zero. Therefore,

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + v 0 y t − 1 2 g t 2 y = y 0 + v 0 y t − 1 2 g t 2 , and solving the quadratic equation for t t .

Because air resistance is negligible, a x = 0 a x = 0 and the horizontal velocity is constant. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t x = x 0 + v x t , where x 0 x 0 is equal to zero

where v x v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0 . v x = v 0 cos θ 0 . Now,

The time t t for both motions is the same, and so x x is

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below, while some of the fragments may now have a velocity in the –x direction due to the forces of the explosion.

[BL] [OL] [AL] Talk about the sample problem. Discuss the variables or unknowns in each part of the problem Ask students which kinematic equations may be best suited to solve the different parts of the problem.

The expression we found for y y while solving part (a) of the previous problem works for any projectile motion problem where air resistance is negligible. Call the maximum height y = h y = h ; then,

This equation defines the maximum height of a projectile . The maximum height depends only on the vertical component of the initial velocity.

Calculating Projectile Motion: Hot Rock Projectile

Suppose a large rock is ejected from a volcano, as illustrated in Figure 5.31 , with a speed of 25.0   m / s 25.0   m / s and at an angle 3 5 ° 3 5 ° above the horizontal. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path.

Breaking this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the time. The time a projectile is in the air depends only on its vertical motion.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

If we take the initial position y 0 y 0 to be zero, then the final position is y = − 20.0  m . y = − 20.0  m . Now the initial vertical velocity is the vertical component of the initial velocity, found from

Substituting known values yields

Rearranging terms gives a quadratic equation in t t

This expression is a quadratic equation of the form a t 2 + b t + c = 0 a t 2 + b t + c = 0 , where the constants are a = 4.90, b = –14.3, and c = –20.0. Its solutions are given by the quadratic formula

This equation yields two solutions t = 3.96 and t = –1.03. You may verify these solutions as an exercise. The time is t = 3.96 s or –1.03 s. The negative value of time implies an event before the start of motion, so we discard it. Therefore,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m / s 14.3 m / s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

Practice Problems

The fact that vertical and horizontal motions are independent of each other lets us predict the range of a projectile. The range is the horizontal distance R traveled by a projectile on level ground, as illustrated in Figure 5.32 . Throughout history, people have been interested in finding the range of projectiles for practical purposes, such as aiming cannons.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 v 0 , the greater the range, as shown in the figure above. The initial angle θ 0 θ 0 also has a dramatic effect on the range. When air resistance is negligible, the range R R of a projectile on level ground is

where v 0 v 0 is the initial speed and θ 0 θ 0 is the initial angle relative to the horizontal. It is important to note that the range doesn’t apply to problems where the initial and final y position are different, or to cases where the object is launched perfectly horizontally.

Virtual Physics

Projectile motion.

In this simulation you will learn about projectile motion by blasting objects out of a cannon. You can choose between objects such as a tank shell, a golf ball or even a Buick. Experiment with changing the angle, initial speed, and mass, and adding in air resistance. Make a game out of this simulation by trying to hit the target.

Check Your Understanding

• Projectile motion is the motion of an object projected into the air and moving under the influence of gravity.
• Projectile motion is the motion of an object projected into the air and moving independently of gravity.
• Projectile motion is the motion of an object projected vertically upward into the air and moving under the influence of gravity.
• Projectile motion is the motion of an object projected horizontally into the air and moving independently of gravity.

What is the force experienced by a projectile after the initial force that launched it into the air in the absence of air resistance?

• The nuclear force
• The gravitational force
• The electromagnetic force
• The contact force

Use the Check Your Understanding questions to assess whether students achieve the learning objectives for this section. If students are struggling with a specific objective, the Check Your Understanding will help identify which objective is causing the problem and direct students to the relevant content.

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Resources Year 12 Physics

Year 12 Physics Practical Investigation | Projectile Motion Experiment

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Sample Physics Practical Assessment Task: Projectile Motion Experiment

Projectile motion experiment is used by most schools for their first Physics practical assessment task. This is because most Projectile Motion practical investigation is relatively easy to design and conduct by students.

A typical Projectile Motion practical assessment task used by schools is outlined below.

In this sample practical assessment task, we are required to investigate the relationship between the range s_x and the launch velocity of a projectile released from an elevated position.

Let’s apply the scientific method to design and conduct a practical investigation for the assessment task outlined above.

Sample Physics Practical Report

The simplest type of projectile motion is a ball being projected horizontally from an elevated position.

In this situation, the range of a projectile is dependent on the time of flight and the horizontal velocity. Hence this experiment is based on the equation s_x=u_xt .

To express the time of flight t in terms of the acceleration due to gravity, we analyse the vertical motion of the projectile

s_y=u_yt+\frac{1}{2}at^2

-h=0+\frac{1}{2}(-g)t^2

t^2=\frac{2h}{g}

t=\sqrt{\frac{2h}{g}}

Hence the range of a projectile can be expressed in terms of the horizontal velocity and the other control variables such as y and g by substituting t=\sqrt{\frac{2h}{g}} expression into s_x=u_xt :

s_x = u_x \times (\sqrt{\frac{2h}{g}})

\therefore s_x = (\sqrt{\frac{2h}{g}}) u_x

2. Variables

Before designing your investigation, all the variables need to be identified.

• Independent variable: Horizontal launch velocity u_x
• Dependent variable: Range \Delta x
• Control variables: Height of the table y , acceleration due to gravity g , the shape of the projectile

Keeping the control variables constant allows the experiment to be more valid.

To learn more about how to improve the validity of your experiment, read the Matrix blog on ‘ Validity, Reliability and Accuracy of Experiments ‘

To determine the relationship between the range of a projectile \Delta x  and its horizontal launch velocity  u_x and use the results to calculate the acceleration due to gravity  g .

• A smooth metal ball is placed at the top of the ramp, and the vertical distance from the ball to the table is measured.
• The ball is rolled down and timed along the 1 \ m horizontal length using a stopwatch. The time is recorded.
• The distance from the foot of the table to its landing point on the carbon paper is observed, measured and recorded.
• Steps 1-4 are then repeated at different heights up the ramp.

The results are given in the table below. Using the times taken for the ball to travel 1 metre. Data collected from the experiment is highlighted in blue.

6. Quantitative Analysis of Results: Graphs and calculations

Calculate the horizontal velocity of the ball as it leaves the table and hence complete the table.

Plot the range of the ball \Delta x against the launch velocity u_x and draw in the line of best fit. 

• The range of the ball is plotted against the horizontal launch velocity.

• A line of best fit is drawn.

Determine the relationship between the launch velocity u_x and the range of the ball \Delta x  and hence discuss its significance

• The relationship between the launch velocity and the range of the ball is linear.  The range of the ball is directly proportional to the horizontal launch velocity: s_x = u_x \times t
• The linear relationship implies that the horizontal launch velocity affects the range but not the time taken to fall from a fixed height. Therefore horizontal and vertical motions are independent of each other.
• This also validates the results expected from the equations of projectile motion.

Use the gradient to find the acceleration due to gravity

7. Qualitative Analysis: Evaluation of method and errors

Let’s investigate the errors, reliability and accuracy of this experiment.

Access our library of Physics Practical Investigations.

Written by DJ Kim

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Physics depth study | the comprehensive guide, year 12 physics textbook review, chapter 1: physics depth study ideas and topics.

What are the possible sources of error in projectile motion experiment?

The main source of error is likely to be the effect of air resistance, which is very difficult to account for theoretically at this level and is systematic in that it should decrease the range of all projectiles launched. Another source of error is likely to be the precision with which the projectile is aimed.

Table of Contents

How do you do a projectile motion experiment?

The simplest type of projectile motion is a ball being projected horizontally from an elevated position. In this situation, the range of a projectile is dependent on the time of flight and the horizontal velocity. Hence this experiment is based on the equation s x = u x t s_x=u_xt sx=uxt.

What is the conclusion of projectile motion lab?

Conclusion: In conclusion, this lab allowed us to investigate projectile motion by determining the initial velocity given to the ball and ultimately predict the range of a projectile. Projectile Motion equations were used to predict the range in this projectile motion lab.

What is the purpose of the projectile motion lab?

The purpose of this experiment is to predict and verify the range and the time-of-flight of a projectile launched at an angle. To predict the range of the projectile when it is shot off a table at some angle above the horizontal, it is necessary first to determine the initial speed ( muzzle velocity ) of the ball.

How do you find the initial velocity of a projectile motion experiment?

What are the sources of error in physics lab?

Common sources of error include instrumental, environmental, procedural, and human. All of these errors can be either random or systematic depending on how they affect the results. Instrumental error happens when the instruments being used are inaccurate, such as a balance that does not work (SF Fig. 1.4).

What are sources of random error?

Sources of random errors natural variations in real world or experimental contexts. imprecise or unreliable measurement instruments . individual differences between participants or units. poorly controlled experimental procedures.

Why is air resistance a source of error?

For example, if you measure gravitational acceleration in a free fall experiment to be larger than 9.81 m/s2, it would be inconsistent to cite air resistance as a source of error (because air resistance would cause the measured acceleration to be less than 9.81 m/s2, not larger).

How did changing the launch angle of a projectile affect its time of flight maximum height and range?

Higher launch angles have higher maximum height The maximum height is determined by the initial vertical velocity. Since steeper launch angles have a larger vertical velocity component, increasing the launch angle increases the maximum height.

What are the three types of projectile motion?

• Oblique projectile motion.
• Horizontal projectile motion.
• Projectile motion on an inclined plane.

What are the characteristics of projectile motion?

Objects experiencing projectile motion have a constant velocity in the horizontal direction, and a constantly changing velocity in the vertical direction. The trajectory resulting from this combination always has the shape of a parabola.

What is the importance of projectile motion in sports?

Understanding how projectile motion works is very beneficial in determining how to best propel an object. For the javelin throw, being able to calculate the different variables helps the athlete to develop a better technique for them personally in order to throw the longest distance.

How do you find final velocity without initial velocity?

• Work out which of the displacement (S), initial velocity (U), acceleration (A) and time (T) you have to solve for final velocity (V).
• If you have U, A and T, use V = U + AT.
• If you have S, U and T, use V = 2(S/T) – U.
• If you have S, A and T, use V = (S/T) + (AT/2).

Can you calculate velocity without time?

Examine the problem to find the displacement of the object and its initial velocity. Plug the acceleration, displacement and initial velocity into this equation: (Final Velocity)^2 = (Initial Velocity) ^2 + 2_(Acceleration)_(Displacement).

How do you find final velocity?

Final Velocity Formula vf=vi+aΔt. For a given initial velocity of an object, you can multiply the acceleration due to a force by the time the force is applied and add it to the initial velocity to get the final velocity.

What are the 3 types of experimental errors?

Three general types of errors occur in lab measurements: random error, systematic error, and gross errors. Random (or indeterminate) errors are caused by uncontrollable fluctuations in variables that affect experimental results.

What percentage of error is acceptable?

In some cases, the measurement may be so difficult that a 10 % error or even higher may be acceptable. In other cases, a 1 % error may be too high. Most high school and introductory university instructors will accept a 5 % error.

How do you explain a mistake in a lab report?

Reread procedures outlined in manuals from before the experiment and your own reflective write up of the experimental steps. Recall the mechanisms you used and any problems that may have come up. This may include measurements in weighing and alterations of steps as necessary. Mark down changes from procedure.

How do you know if a error is systematic or random?

Random error causes one measurement to differ slightly from the next. It comes from unpredictable changes during an experiment. Systematic error always affects measurements the same amount or by the same proportion, provided that a reading is taken the same way each time. It is predictable.

What is a zero error in physics?

zero error Any indication that a measuring system gives a false reading when the true value of a measured quantity is zero, eg the needle on an ammeter failing to return to zero when no current flows. A zero error may result in a systematic uncertainty.

How do you reduce random and systematic errors?

Calibration, when feasible, is the most reliable way to reduce systematic errors . To calibrate your experimental procedure, you perform it upon a reference quantity for which the correct result is already known.

Do heavier objects fall faster with air resistance?

Galileo discovered that objects that are more dense, or have more mass, fall at a faster rate than less dense objects, due to this air resistance.

Why does mass not affect projectile motion?

When writing equations of motion for a dropped object, mass is in the equations in 2 places and they cancel out. That is basically the reason that mass does not affect the results of analysis of a projectile.

Why do we neglect air resistance in projectile motion?

As an object travels through the air, it encounters a frictional force that slows its motion called air resistance. Air resistance does significantly alter trajectory motion, but due to the difficulty in calculation, it is ignored in introductory physics .

Which factor is most important in projectile motion?

Complete answer: A projectile is an object in which the only force acting on an object is gravity. There are some examples of projectiles. When An object is dropped from its rest position it is known as a projectile which ensures that the influence of air resistance is negligible.

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Error propagation in projectile motion lab

A lab activity designed for introductory physics students to compare measured and calculated ranges for a projectile launched from a reference height is presented.

A lab activity designed for students to compare measured and calculated ranges for a projectile launched from a reference height is presented here. Students used statistical and error propagation techniques to analytically determine the error bounds associated with measured and calculated projectile ranges as well as t-statistics to determine how well the measured and calculated ranges agreed. For all launch angles used in this work, 90% of students found that there was no statistically significant difference ( p < 1 ) (p < 1) ( p < 1 ) between the measured and calculated ranges.

Introduction

Every physical measurement has an associated error. Including error in reported physical results establishes a basis for deciding whether a scientific hypothesis should be accepted or rejected. As such, reporting errors establishes a level of confidence associated with the measured value, reflects the quality of the experiment, and allows for comparison with theoretical values. Apart from reporting the errors, understanding error propagation is important as it adds validation to reported experimental results (Chhetri, 2013; Baird, 1995; Labs for College Physics, n.d.; Monteiro et al., 2021). The error propagation technique is a skill that the American Association of Physics Teachers (AAPT) recommends that all Physics students develop (AAPT, 2014). Experimental data are routinely used to make conclusions in physics, astronomy, chemistry, life sciences, engineering and many other fields of study.  Error propagation techniques are particularly helpful to students or experimentalists in these fields to apply best practices to make estimates for experimental measures and report inaccuracies within practical reasons. (Lippmann, 2003 & Berendsen, 2011). However, error propagation techniques are rarely emphasized in introductory physics classes, as such, few undergraduate lab activities exist that focus on these techniques (Taylor, 1985; Allen, 2021). In fact, error propagation may only be included in junior/senior level physics classes and are largely omitted or greatly simplified in introductory classes (Faux & Godolphin, 2019; Purcell, 1974).

This paper presents a lab activity designed to introduce students to error propagation methods as applied to the range measurements of a projectile. Students measured ranges of projectile launched at various angles from a reference height and compared their measured results to calculated ranges by employing statistical and error propagation techniques. Students then used t-statistics to establish the level of confidence in their measurements. This lab activity was implemented in an introductory physics course at Georgia Gwinnett College (GGC). Georgia Gwinnett College is an access, four-year, minority-serving, Hispanic-serving, liberal arts, public institution committed to student success. The college emphasizes an integrated educational experience for students and encourages new teaching pedagogy as well as the innovative use of technology. Physics class sizes at GGC are limited to a maximum of 24 students and taught in a studio-style setting in 2 hours and 45-minute sections.

The experimental data collection for this activity took 45 minutes and the remainder of the class time was spent on data analysis and completing the lab report. The authors intend that upon completion of this activity, students understand how to: 1) calculate averages and standard errors from experimental data, 2) apply error propagation techniques, 3) report errors in measurements, and 4) use t-tests to compare the measured and calculated results. Typically, GGC students enrolled in introductory calculus-based physics classes are pre-engineering majors and have only completed Calculus I and have not necessarily taken a statistics course.

The purpose of this lab activity is to introduce students in introductory physics courses how to apply error propagation techniques in establishing error bounds in experimental measurements.

Learning objectives

Typically, most students enrolling in introductory calculus-based physics classes have only completed calculus I and have not necessarily taken a statistics course. The authors intend that upon completion of this activity, students would be able to understand how to 1) find averages and standard errors from experimental data, 2) report errors in measurements, 3) apply error propagation techniques, and 4) use t-tests to compare the measured and calculated results.

Materials and Equipment

The materials required for this lab include a PASCO projectile launcher fitted with a protractor (ME-6800), small metal PASCO launch balls (ME-9859), meter stick, carbon paper, and safety glasses.

Prior to the start of the experiment, the concepts of projectile motion were reviewed. This review with students allowed for further questions, discussions, and improved understanding of projectile motion. The overviews shared with students are presented in the following two sections (Overview 1 and Overview 2).

Figure 1 shows the variables associated with a projectile launched horizontally from a known height with velocity ( v o ) v_{o}) v o ​ ) .

FIG. 1. Projectile launched horizontally ( θ = 0 ) (\theta = 0) ( θ = 0 ) , with velocity ( v o ) v_{o}) v o ​ ) from a known launch height (Δy = -h) and the corresponding range ( Δ x \mathrm{\Delta}x Δ x ).

For such a horizontal launch, Δ x \mathrm{\Delta}x Δ x is the horizontal distance (range) covered by the projectile, Δ y \mathrm{\Delta}y Δ y the known launch height, v o x v_{ox} v o x ​ and v o y v_{oy} v oy ​ are the x- and y-components of initial launch velocity respectively, and t t t is the time of flight. When kinematic equations are applied to projectile motion with constant downward acceleration (g= 9.81 m/s 2 ), the expressions for Δ x \mathrm{\Delta}x Δ x and Δ y \mathrm{\Delta}y Δ y become:

The time of flight is obtained from equation (2) as:

The expression for v o v_{o} v o ​ can be obtained by combining equations (1) and (3):

Figure 2 shows the trajectory of a projectile launched at an angle ( θ = 0 \mathrm{\theta=0} θ = 0 ) with respect to the horizontal, from a reference height ( Δ y = − h \mathrm{\Delta}y=-h Δ y = − h ). In this experiment, air resistance is considered to be negligible and g=9.80 m/s 2 .

Figure 2: Projectile launched at a known angle, from a known launch height (-h) showing the corresponding range to be measured.

The kinematics equations in the y-direction can be written:

  v y = v 0 y − g t v_{y} = v_{0y} - gt v y ​ = v 0 y ​ − g t (5)

Δ y = h + v 0 y t − 1 2 g t 2 \mathrm{\Delta}y = h + v_{0y}t - \frac{1}{2}gt^{2} Δ y = h + v 0 y ​ t − 2 1 ​ g t 2 (6)

v y 2 = v 0 y 2 − 2 g ( y − h ) v_{y}^2 = v_{0y}^2 - 2g(y-h) v y 2 ​ = v 0 y 2 ​ − 2 g ( y − h ) (7)

Where   v 0 y = v 0 sin ⁡ ( θ ) = 0 {v_{0y} = v_{0}\sin(\theta)=0} v 0 y ​ = v 0 ​ sin ( θ ) = 0 and v y v_{y} v y ​  is the y-components of the projectile’s velocity at any point along the trajectory.

Additionally, for conciseness, the basic statistical formulas needed to calculate the average and standard errors of measurements, t-statistics for comparing two values, and error propagation rules are summarized in Appendix 1. Appendix 2 is provided to show the derivation of the equations used in calculating the errors associated with launch velocity, time of flight, and calculated range ( δ v o ,    δ t o ,    a n d   δ ( Δ x c a l ) \delta v_{o},\ \ \delta t_{o},\ \ and\ \delta(\mathrm{\Delta}x_{cal}) δ v o ​ ,     δ t o ​ ,     an d   δ ( Δ x c a l ​ ) ), respectively.v

The lab activity is comprised of four parts:

Part I: Determine the initial launch velocity, v o v_{o} v o ​ , using measured average range, ( Δ x ‾ ) \left( \overline{\mathrm{\Delta}x} \right) ( Δ x ) , from horizontal launch and known launch height, Δ y \mathrm{\Delta}y Δ y .

Part II: Measure projectile range ( Δ x m e a s {\mathrm{\Delta}x}_{meas} Δ x m e a s ​ ) launched at various angles (35 o , 45 o , & 55 o ), using calculated initial velocity ( v o ) v_{o}) v o ​ ) with associated errors ( δ ( v o ) \delta(v_{o}) δ ( v o ​ ) ) from Part 1.

Part III: Calculate ranges ( Δ x c a l {\mathrm{\Delta}x}_{cal} Δ x c a l ​ ) with associated error ( δ ( Δ x c a l c ) \delta(\mathrm{\Delta}x_{calc}) δ ( Δ x c a l c ​ ) ) for projectiles launched at different angles (35 o , 45 o , & 55 o )

Part IV: Compare measured ranges ( Δx meas ) to the calculated ranges ( Δ x c a l {\mathrm{\Delta}x}_{cal} Δ x c a l ​ ) using t-statistics.

Part I: Determine initial launch velocity, v o \mathbf{v}_{\mathbf{o}} v o ​ using measured range from horizontal launch

A projectile was launched from a table onto the floor as shown in the experimental setup in Figure 1. Using a meter-stick with a 0.05 cm accuracy, the range ( Δ x ) \mathrm{\Delta}x) Δ x ) and reference height ( Δ y ) \mathrm{\Delta}y) Δ y ) were recorded. The range measurements were repeated 10 times, keeping Δ y \mathrm{\Delta}y Δ y the same. The average range, ( Δ x ‾ ) , \left( \overline{\mathrm{\Delta}x} \right), ( Δ x ) , and corresponding error, δ ( Δ x ‾ ) , \delta\left( \overline{\mathrm{\Delta}x} \right), δ ( Δ x ) , were then calculated using equations A1.1 through A1.3, from Appendix 1. Table 1 shows a sample of student data; values of Δ x ‾ ,   δ ( Δ x ‾ ) ,   Δ y ,   a n d   δ ( Δ y ) \overline{\mathrm{\Delta}x},\ \delta\left( \overline{\mathrm{\Delta}x} \right),\ \mathrm{\Delta}y,\ and\ \delta(\mathrm{\Delta}y) Δ x ,   δ ( Δ x ) ,   Δ y ,   an d   δ ( Δ y ) are shown. These measurements were used to calculate the initial launch velocity using equation 4.

Part II: Measuring projectile range ( Δ x m e a s \mathbf{\mathrm{\Delta}x}_{\mathbf{meas}} Δ x meas ​ ) launched at various angles (35 o , 45 o , & 55 o )

Following the experimental set-up shown in Figure 2, a projectile was launched at three different angles and Δ x m e a s ( θ ) {\mathrm{\Delta}x}_{meas}(\theta) Δ x m e a s ​ ( θ ) was measured for each. These Δ x m e a s {\mathrm{\Delta}x}_{meas} Δ x m e a s ​ measurements were repeated 10 times for each angle. Table 2 shows sample of student data listing the values of ∆y, ( Δ x ‾ m e a s ) {\overline{\mathrm{\Delta}x}}_{meas}) Δ x m e a s ​ ) , and ( δ ( Δ x ‾ ) \delta(\overline{\mathrm{\Delta}x}) δ ( Δ x ) ) for the different launch angles. Here, Δ x ‾ m e a s {\overline{\mathrm{\Delta}x}}_{meas} Δ x m e a s ​ , and ( δ Δ x ‾ m e a s ) \delta{\overline{\mathrm{\Delta}x}}_{meas}) δ Δ x m e a s ​ ) denotes the average of the 10 values of Δ x m e a s {\mathrm{\Delta}x}_{meas} Δ x m e a s ​ and associated errors, respectively.

Part III: Calculating ranges ( Δ x c a l \mathbf{\mathrm{\Delta}x}_{\mathbf{cal}} Δ x cal ​ ), for projectiles launched at different angles (35 o , 45 o , & 55 o )

This part dealt with calculated ranges, Δ x c a l {\mathrm{\Delta}x}_{cal} Δ x c a l ​ , for the projectile launched at the different angles using the initial velocity v o ± δ v o = ( 2.835 ± 0.004 ) m s v_{o} \pm \delta v_{o} = (2.835 \pm 0.004)\frac{m}{s} v o ​ ± δ v o ​ = ( 2.835 ± 0.004 ) s m ​ and the known launch heights from Part I. To calculate Δ x c a l {\mathrm{\Delta}x}_{cal} Δ x c a l ​ , students

Determined the initial x- and y-components of the launch velocities ( v o x   a n d   v o y ) v_{ox}\ and\ v_{oy}) v o x ​   an d   v oy ​ ) ,

Found the time of flight ( t c a l ) t_{cal}) t c a l ​ ) of the projectile,

Calculated, Δ x c a l = v o x t c a l {\mathrm{\Delta}x}_{cal} = v_{ox} t_{cal} Δ x c a l ​ = v o x ​ t c a l ​ ,

and recorded the calculated range in the form:

Δ x c a l + δ ( Δ x c a l ) {\mathrm{\Delta}x}_{cal} + {\delta(\mathrm{\Delta}x}_{cal}) Δ x c a l ​ + δ ( Δ x c a l ​ ) (8)

Table 3 lists the calculated values of v o x v_{ox} v o x ​ and v o y v_{oy} v oy ​ for the different launch angles along with corresponding errors δ ( v o x ) \delta (v_{ox}) δ ( v o x ​ ) and δ ( v o y ) \delta (v_{oy}) δ ( v oy ​ ) using equations A2.6 and A2.7. Table 4 lists the time of flight t c a l t_{cal} t c a l ​ and associated error δ ( t c a l ) \delta (t_{cal}) δ ( t c a l ​ ) for the different angles using equations A2.8 and A2.12. Finally, Table 5 lists the calculated range, Δ x c a l ( θ ) \mathrm{\Delta}x_{cal}(\theta) Δ x c a l ​ ( θ ) , for projectile at different launch angles.

Data Collection and Analysis

The data presented in this section represents the measured projectile ranges launched horizontally (0.0 o ) and ranges for projectile launched at different angles (35.0 o , 45.0 o , and 55.0 o ). Also, the calculated projectile ranges and initial launch velocities at the different launch angles are presented. Furthermore, the calculated projectile time of flight, with associated error is presented. Finally, an analysis of the measured range is compared to the calculated range using t-statistics. The provided data are complete and match the descriptions in the contribution. Readers who are interested in obtaining a copy of the lab materials provided to students should contact Joseph Ametepe ( [email protected] ).

TABLE 1: Measured ranges for horizontal launch

The above measured values of Δ x ‾ ± δ ( Δ x ‾ ) \overline{\mathrm{\Delta}x} \pm \delta(\overline{\mathrm{\Delta}x}) Δ x ± δ ( Δ x ) and Δ y ‾ ± δ ( Δ y ‾ ) \overline{\mathrm{\Delta}y} \pm \delta(\overline{\mathrm{\Delta}y}) Δ y ​ ± δ ( Δ y ​ ) were used to calculate v o v_{o} v o ​ along with its error δ ( v o ) \delta(v_{o}) δ ( v o ​ ) by using equations (4) and (A2.5) and the result is reported as v o ± δ v o = ( 2.835 ± 0.004 ) m s . v_{o} \pm \delta v_{o} = (2.835 \pm 0.004)\frac{m}{s}. v o ​ ± δ v o ​ = ( 2.835 ± 0.004 ) s m ​ .

TABLE 2: Measured ranges for projectile launched at different angles (precision of compass δ θ = 0.5 o \delta\theta = {0.5}^{o} δ θ = 0.5 o )

It should be noted that the error in the 55-degree measurement is significantly higher than for the other two angles. This is at least partially due to the fact that the work done by the spring has been neglected in our calculations and here it is non-negligible (Schnick, 1994).

Table 3 shows a sample of student data listing the calculated values of v o x   a n d   v o y v_{ox}\ and\ v_{oy} v o x ​   an d   v oy ​ for the different launch angles with associated errors δ ( v o x ) \delta\left( v_{ox} \right) δ ( v o x ​ ) and δ ( v o y ) \delta\left( v_{oy} \right) δ ( v oy ​ ) using A2.6 and A2.7, derived in Appendix 2.

δ ( v o x ) = ∣ v o c o s θ ∣ ( δ ( v o ) v o ) 2 + ( ( s i n θ ) δ θ ) cos ⁡ θ ) 2 \delta\left( v_{ox} \right) = \left| v_{o}{cos}\theta \right|\sqrt{\left( \frac{\delta\left( v_{o} \right)}{v_{o}} \right)^{2} + \left( \frac{{(sin}{\theta)} \delta\theta)}{\cos\theta} \right)^{2}} δ ( v o x ​ ) = ∣ v o ​ cos θ ∣ ( v o ​ δ ( v o ​ ) ​ ) 2 + ( c o s θ ( s in θ ) δ θ ) ​ ) 2 ​ (A2.6)

δ ( v o y ) = ∣ v o s i n θ ∣ ( δ ( v o ) v o ) 2 + ( ( c o s   θ ) δ θ sin ⁡ θ ) 2 \delta\left( v_{oy} \right) = \left| v_{o}{sin}\theta \right|\sqrt{\left( \frac{\delta\left( v_{o} \right)}{v_{o}} \right)^{2} + \left( \frac{({cos\ \theta)}{\delta\theta}}{\sin\theta} \right)^{2}} δ ( v oy ​ ) = ∣ v o ​ s in θ ∣ ( v o ​ δ ( v o ​ ) ​ ) 2 + ( s i n θ ( cos   θ ) δ θ ​ ) 2 ​ (A2.7)

TABLE 3: Initial launch velocities ( v o x   a n d   v o y ) v_{ox}\ and\ v_{oy}) v o x ​   an d   v oy ​ ) for different launch angles ( θ ) \theta) θ )

Table 4 lists the time of flight t c a l t_{cal} t c a l ​ and associated error δ ( t c a l ) \delta\left( t_{cal} \right) δ ( t c a l ​ ) for the different angles using A2.8 and A2.12 derived in Appendix 2, reported below.

t c a l = ( v o sin ⁡ θ ±   ( v o sin ⁡ θ ) 2 + 2 g h   ) g t_{cal} = \frac{\left( v_{o} \sin{\theta \pm}\ \sqrt{\left( v_{o} \sin\theta \right)^{2} + 2 g h}\ \right)}{g} t c a l ​ = g ( v o ​ s i n θ ±   ( v o ​ s i n θ ) 2 + 2 g h ​   ) ​ (A2.8)

δ t c a l = 1 g { v o sin ⁡ θ ( δ v o v o ) 2 + ( cot ⁡ θ δ θ ) 2 + ( ( s i n θ ) δ v o ) 2 ( v o sin ⁡ θ ) 2 + 2 g h + ( v o ( cos ⁡ θ ) δ θ ) 2 ( v o sin ⁡ θ ) 2 + 2 g h } {\delta t}_{cal} = \frac{1}{g}\left\{ v_{o} \sin\theta\sqrt{\left( \frac{\delta v_{o}}{v_{o}} \right)^{2} + {(\cot{\theta \delta\theta)}}^{2} + \frac{{{((sin}{\theta)} \delta v_{o})}^{2}}{{(v_{o} \sin{\theta)}}^{2} + 2gh} + \frac{\left( v_{o}\left( \cos\theta \right) \delta\theta \right)^{2}}{{(v_{o} \sin{\theta)}}^{2} + 2gh}} \right\} δ t c a l ​ = g 1 ​ { v o ​ sin θ ( v o ​ δ v o ​ ​ ) 2 + ( cot θ δ θ ) 2 + ( v o ​ s i n θ ) 2 + 2 g h (( s in θ ) δ v o ​ ) 2 ​ + ( v o ​ s i n θ ) 2 + 2 g h ( v o ​ ( c o s θ ) δ θ ) 2 ​ ​ } (A2.12)

TABLE 4: Calculated t c a l   a n d ,    δ t c a l t_{cal}\ and,\ \ \delta t_{cal} t c a l ​   an d ,     δ t c a l ​ for different launch angles

Table 5 lists the calculated range values ( Δ x c a l = v o x t c a l {\mathrm{\Delta}x}_{cal} = v_{ox} t_{cal} Δ x c a l ​ = v o x ​ t c a l ​ ) and associated errors δ ( Δ x c a l ) \delta(\mathrm{\Delta}x_{cal}) δ ( Δ x c a l ​ ) using the values from Tables 4 and 5and equation A2.13 from Appendix 2 for different angles and reported below.

δ ( Δ x c a l ) = ∣ Δ x c a l ∣ ( δ v 0 x v o x ) 2 + ( δ t c a l t c a l ) 2 \delta\left( {\mathrm{\Delta}x}_{cal} \right) = \left| {\mathrm{\Delta}x}_{cal} \right| \sqrt{\left( \frac{\delta v_{0x}}{v_{ox}} \right)^{2}{+ \left( \frac{\delta t_{cal}}{t_{cal}} \right)}^{2}} δ ( Δ x c a l ​ ) = ∣ Δ x c a l ​ ∣ ( v o x ​ δ v 0 x ​ ​ ) 2 + ( t c a l ​ δ t c a l ​ ​ ) 2 ​ (A2.13)

TABLE 5: Calculated range, Δ x c a l ( θ {\mathrm{\Delta}x}_{cal}(\theta Δ x c a l ​ ( θ ) for projectile at different launch angles

TABLE 6: Measured and calculated ranges, and p p p values

Part IV: Analysis of measured and calculated ranges ( Δ x m e a s   a n d   Δ x c a l \mathbf{\mathrm{\Delta}x}_{\mathbf{meas}}\mathbf{\ and\ }\mathbf{\mathrm{\Delta}x}_{\mathbf{cal}} Δ x meas ​   and   Δ x cal ​ )

in this part, students compared Δ x m e a s   a n d   Δ x c a l {\mathrm{\Delta}x}_{meas}\ and\ {\mathrm{\Delta}x}_{cal} Δ x m e a s ​   an d   Δ x c a l ​ by employing t-statistics. In this section, Δ x m − c {\mathrm{\Delta}x}_{m - c} Δ x m − c ​ is used to denote the absolute value of the difference between the measured and calculated range with corresponding error as δ ( Δ x m − c ) {\mathrm{\delta}({\Delta}x}_{m - c}) δ ( Δ x m − c ​ ) . The p-value is then given as p = Δ x m − c δ ( Δ x m − c ) p = \frac{{\mathrm{\Delta}x}_{m - c}}{\delta\left( {\mathrm{\Delta}x}_{m - c} \right)} p = δ ( Δ x m − c ​ ) Δ x m − c ​ ​ , where

Δ x m − c =   ∣ Δ x m e a s − Δ x c a l ∣ {\mathrm{\Delta}x}_{m - c} = \ \left| {\mathrm{\Delta}x}_{meas} - {\mathrm{\Delta}x}_{cal} \right| Δ x m − c ​ =   ∣ Δ x m e a s ​ − Δ x c a l ​ ∣ (A1.6)

δ ( Δ x m − c ) = ( δ Δ x m e a s ) 2 + ( δ Δ x c a l ) 2 {\mathrm{\delta}({\Delta}x}_{m - c})=\sqrt{\left( \delta{\mathrm{\Delta}x}_{meas} \right)^{2} + \left( \delta{\mathrm{\Delta}x}_{cal} \right)^{2}} δ ( Δ x m − c ​ ) = ( δ Δ x m e a s ​ ) 2 + ( δ Δ x c a l ​ ) 2 ​ (A1.7)

Table 6 shows measured and calculated ranges from Tables 3 and 4, and p p p values.

Data in Table 2 clearly show that the lunch angle of θ = 4 5 o \theta = 45^{o} θ = 4 5 o does not yield the maximum range. At this point, students were directed to explore, discuss among themselves, and hypothesize why the launch angle of θ = 4 5 o \theta = 45^{o} θ = 4 5 o did not yield the maximum range in this case where the projectile is launched from an initial launch height . This piece of the exercise is explicitly included for students to distinguish between a projectile launched from ground level versus one launched from an initial launch height.

Students were encouraged to explore the meaning of their t-statistics values, especially the ratio of ∣ Δ x m e a s − Δ x c a l ∣ \left| {\mathrm{\Delta}x}_{meas} - {\mathrm{\Delta}x}_{cal} \right| ∣ Δ x m e a s ​ − Δ x c a l ​ ∣ to ( δ Δ x m e a s ) 2 + ( δ Δ x c a l ) 2 \sqrt{\left( \delta{\mathrm{\Delta}x}_{meas} \right)^{2} + \left( \delta{\mathrm{\Delta}x}_{cal} \right)^{2}} ( δ Δ x m e a s ​ ) 2 + ( δ Δ x c a l ​ ) 2 ​ and relate their results to Table 7 in Appendix 1. Students noted that p < 1 p < 1 p < 1 indicates no statistical difference between measured and calculated ranges.

It should be noted that students can complete many of the calculations presented in this work by hand, however, they can also use other programs such as Mathcad or Excel (Gardenier et al., 2011; Donato & Metz, 1988; de Levie, 2000).

Conclusions

In Part I of this activity, students investigated a projectile fired horizontally from a reference height. With their recorded range and launch height, they were able to calculate the initial launch velocity, v o v_{o} v o ​ with associated error, δ v o \delta v_{o} δ v o ​ . In Part II, students launched the projectile at various angles, to measure ranges, Δ x m e a s \mathrm{\Delta}x_{meas} Δ x m e a s ​ . In Part III, students calculated time of flight, t t t with corresponding errors, δ t \delta t δ t , calculated range, Δ x c a l \mathrm{\Delta}x_{cal} Δ x c a l ​ , with associated δ ( Δ x c a l ) \delta(\mathrm{\Delta}x_{cal}) δ ( Δ x c a l ​ ) . In Part IV, students compared the measured and calculated values of Δ x \mathrm{\Delta}x Δ x , by using t-statistics. Students reported p-values using equation 9:

p = ∣ Δ x m e a s − Δ x c a l c ∣ ( δ Δ x m e a s ) 2 + ( δ Δ x c a l ) 2 p = \frac{\left| {\mathrm{\Delta}x}_{meas} - {\mathrm{\Delta}x}_{calc} \right|}{\sqrt{\left( \delta{\mathrm{\Delta}x}_{meas} \right)^{2} + \left( \delta{\mathrm{\Delta}x}_{cal} \right)^{2}}} p = ( δ Δ x m e a s ​ ) 2 + ( δ Δ x c a l ​ ) 2 ​ ∣ Δ x m e a s ​ − Δ x c a l c ​ ∣ ​ (9)

Student data for launch angles 35 o , 45 o , and 55 o is given in Table 6.

For all launch angles, 90% of students reported that there was no significant difference between the measured and calculated ranges. This was valid within the error bounds established by the error propagation techniques employed.

Through this activity, our students learned transferable skills of how to calculate averages and standard errors from repeated measurements and how to use t-statistics to compare measured and calculated values. Furthermore, students learned the condition of maximum projectile range occurring at 45 degrees applies only when the projectile is fired from ground level.

Reflection & Moving Forward

Completing this activity presented students with transferable skills and learning opportunities associated with experimental measurements, applying error propagation, theoretical calculations of ranges for projectile launched from a reference height, and applying t-statistics to compare two values.

Specifically, students understood (from the set up in Figure 1) that the initial launch velocity can be determined from range and launch height measurements. Furthermore, students were able to apply error propagation techniques to determine error-associated boundaries for their projectile range measurements and calculations. The error propagation exercise portion was an important component of this activity as students do not traditionally cover error propagation in many introductory lab activities (Taylor, 1985; Allen, 2021; Faux & Godolphin, 2019; Purcell, 1974).

Another learning opportunity was associated with the use of t-statistics to compare the measured and calculated ranges. For all launch angles (35 o , 45 o , and 55 o ) used in this work, 90% of students found that there was no statistically significant difference ( p < 1 ) (p < 1) ( p < 1 ) between the measured and calculated ranges. Therefore, their measured values were acceptable within their error bounds. Introducing this concept in an introductory course gives students an analytical tool to use in order to critically evaluate their results rather than simply relying on a “gut feeling” that their results are “good” or “bad.”

The results of the small group of students who had statistically significant differences between calculated and measured range values were investigated. A common error that occurs when students perform projectile motion labs is the failure to ensure that the exit of the launcher barrel is consistently set as the origin of measurements. This critical issue led to values outside the error bounds. If the barrel extends beyond the origin of measurement, errors due to a subtle conservation of energy considerations that needs to be compensated for are introduced. Compensating for these errors will be the focus of a forthcoming paper.

Appendix 1: Review of Statistics and Error Propagation

This Appendix summarizes simple statistics and error propagation methods.

Review of Statistics

Presented here is a review of finding the mean, standard deviation and associated errors. Suppose we measure a quantity N times (N ≥ 10), then

The mean ( x ‾ ) \overline{x}) x ) of the N measurements can be written as:

x   ‾ = ∑ 1 N x i N \overline{x\ } = \frac{\sum_{1}^{N}x_{i}}{N} x   = N ∑ 1 N ​ x i ​ ​ (A1.1)

The standard deviation σ of the sample measurements can be written as:

σ = ( ∑ 1 N ( x i − x ‾ ) 2 ) / ( N − 1 )   \sigma = \sqrt{(\sum_{1}^{N}{{(x_{i} - \overline{x})}^{2})/(N - 1)\ }} σ = ( ∑ 1 N ​ ( x i ​ − x ) 2 ) / ( N − 1 )   ​ (A1.2)

The “error” in the measurement is the standard error ( σ N = σ / N ) \sigma_{N} = \sigma/\sqrt{N}) σ N ​ = σ / N ​ ) can be written as:

σ N = σ N \sigma_{N} = \frac{\sigma}{\sqrt{N}} σ N ​ = N ​ σ ​ (A1.3)

The mean value of measurement with error is reported as:

M e a s u r e m e n t =   x   ‾ ±   σ N Measurement = \ \overline{x\ } \pm {\ \sigma}_{N} M e a s u re m e n t =   x   ±   σ N ​ (A1.4)

The t-statistics to determine whether the difference between two independent values x ‾ 1   &   x ‾ 2 {\overline{x}}_{1}\ \&\ {\overline{x}}_{2} x 1 ​   &   x 2 ​ is significant or not can be determined from:

p = Δ x ‾ σ Δ x p = \frac{\overline{\mathrm{\Delta}x}}{\sigma_{\mathrm{\Delta}x}} p = σ Δ x ​ Δ x ​ (A1.5)

Δ x ‾ = ∣ x ‾ 1 − x ‾ 2   ∣ \overline{\mathrm{\Delta}x} = \left| {\overline{x}}_{1} - {\overline{x}}_{2}\ \right| Δ x = ∣ x 1 ​ − x 2 ​   ∣ (A1.6)

σ Δ x = σ x 1 2 + σ x 2 2 \sigma_{\mathrm{\Delta}x} = \sqrt{\sigma_{x1}^{2} + \sigma_{x2}^{2}} σ Δ x ​ = σ x 1 2 ​ + σ x 2 2 ​ ​ (A1.7)

where σ x 1 \sigma_{x1} σ x 1 ​ and σ x 2 \sigma_{x2} σ x 2 ​ are the standard errors associated with x ‾ 1 {\overline{x}}_{1} x 1 ​ and x ‾ 2 {\overline{x}}_{2} x 2 ​ , respectively.

TABLE 7: Meaning of p values

Rules for Error Propagation

When dealing with error propagation, there are specific rules (Harvey, 2009; Allen, 2021) to be followed that are summarized in A1.8 – A1.14. In this activity, we assumed that the variables are uncorrelated and, therefore, the error is evaluated using simple sums of partial derivatives (Taylor, 1997). If X ,   Y ,   Z ,   … X,\ Y,\ Z,\ \ldots X ,   Y ,   Z ,   … are measured values with δ X ,   δ Y ,   δ Z , … . \delta X,\ \delta Y,\ \delta Z,\ldots. δ X ,   δ Y ,   δ Z , … . as associated errors, then the following equations apply:

R   =   X   ±   Y ;           δ R   =   ( δ X ) 2 + ( δ Y ) 2 R\ = \ X\ \pm \ Y;\ \ \ \ \ \ \ \ \ \delta R\ = \ \sqrt{{(\delta X)}^{2} + ({\delta Y)}^{2}} R   =   X   ±   Y ;                   δ R   =   ( δ X ) 2 + ( δ Y ) 2 ​ (A1.8)

Multiplication

R   =   X    Y ;    δ R   = ∣ R ∣ ( δ X X ) 2 + ( δ Y Y ) 2 R\ = \ X\ \ Y;\ \ \delta R\ = |R|\sqrt{{(\frac{\delta X}{X})}^{2} + {(\frac{\delta Y}{Y})}^{2}} R   =   X     Y ;     δ R   = ∣ R ∣ ( X δ X ​ ) 2 + ( Y δ Y ​ ) 2 ​ (A1.9)

R   = X Y   ;            δ R   =   ∣ R ∣ ( δ X X ) 2 + ( δ Y Y ) 2 R\ = \frac{X}{Y}\ ;\ \ \ \ \ \ \ \ \ \ \delta R\ = \ |R|\sqrt{{\left( \frac{\delta X}{X} \right)^{2} + \left( \frac{\delta Y}{Y} \right)^{2}}^{}} R   = Y X ​   ;                     δ R   =   ∣ R ∣ ( X δ X ​ ) 2 + ( Y δ Y ​ ) 2 ​ (A1.10)

Multiplying by a constant, c c c

R = c X ;   δ R   = ∣ c ∣ δ X R = c X;\ \delta R\ = |c| \delta X R = c X ;   δ R   = ∣ c ∣ δ X (A1.11)

Single-valued functions: For a single-valued function R(X), the associated error δ ( R ( x ) ) \delta(R(x)) δ ( R ( x )) is:

δ ( R ( X ) ) = ( d R ( X ) d X ) δ X \delta\left( R(X) \right) = \left( \frac{dR(X)}{dX} \right)\delta X δ ( R ( X ) ) = ( d X d R ( X ) ​ ) δ X (A1.12)

Multi-valued Functions: For a multi-valued function R(X, Y, Z, …), in which X, Y, Z, … are uncorrelated, then:

R = R ( X , Y ,   … . ) ;                  δ R = ( ∂ R ∂ X δ X ) 2 + ( ∂ R ∂ Y δ Y ) 2 + … R = R(X,Y,\ \ldots.);\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta R = \sqrt{{(\frac{\partial R}{\partial X} \delta X)}^{2} + {(\frac{\partial R}{\partial Y} \delta Y)}^{2} + \ldots} R = R ( X , Y ,   … . ) ;                                 δ R = ( ∂ X ∂ R ​ δ X ) 2 + ( ∂ Y ∂ R ​ δ Y ) 2 + … ​ (A1.13)

Polynomial Functions (R based on polynomial function of one variable X)

R = X n ;    δ R =   ∣ n ∣ X n − 1 δ X   o r     δ R =   ∣ n ∣ δ X ∣ X ∣ ∣ R ∣ R = X^{n};\ \ \delta R = \ |n|X^{n - 1} \delta X\ \mathbf{or\ }\ \delta R = \ |n| \frac{\delta X}{|X|} |R| R = X n ;     δ R =   ∣ n ∣ X n − 1 δ X   or     δ R =   ∣ n ∣ ∣ X ∣ δ X ​ ∣ R ∣ (A1.14)

Appendix 2: Derivations of Equations with their corresponding error

The initial launch velocity and associated error is derived by starting with equation (4). v o v_{o} v o ​ can be written as:

  v o = g 2 ( Δ x 1 Δ y ) \ v_{o} = \sqrt{\frac{g}{2}} (\mathrm{\Delta}x \frac{1}{\sqrt{\mathrm{\Delta}y}})   v o ​ = 2 g ​ ​ ( Δ x Δ y ​ 1 ​ ) (A2.1)

δ v o =   g 2 δ F ( Δ x , Δ y ) \delta v_{o} = \ \sqrt{\frac{g}{2}}\delta F(\mathrm{\Delta}x,\mathrm{\Delta}y) δ v o ​ =   2 g ​ ​ δ F ( Δ x , Δ y ) (A2.2)

F ( Δ x , Δ y ) = ( Δ x ) ( 1 Δ y ) ) F(\mathrm{\Delta}x,\mathrm{\Delta}y) = (\mathrm{\Delta}x) (\frac{1}{\sqrt{\mathrm{\Delta}y}})) F ( Δ x , Δ y ) = ( Δ x ) ( Δ y ​ 1 ​ )) (A2.3)

Using equation (A1.9), δ F ( Δ x , Δ y ) = ( ∂ F ∂ Δ x ) 2 ( δ Δ x ) 2 + ( ∂ F ∂ Δ y ) 2 ( δ Δ y ) 2 \delta F(\mathrm{\Delta}x,\mathrm{\Delta}y) = \sqrt{\left( \frac{\partial F}{\partial\mathrm{\Delta}x} \right)^{2} (\delta\mathrm{\Delta}x)^{2} + \left( \frac{\partial F}{\partial\mathrm{\Delta}y} \right)^{2} (\delta\mathrm{\Delta}y)^{2}} δ F ( Δ x , Δ y ) = ( ∂ Δ x ∂ F ​ ) 2 ( δ Δ x ) 2 + ( ∂ Δ y ∂ F ​ ) 2 ( δ Δ y ) 2 ​

with ∂ F ∂ ( Δ x ) = ( 1 Δ y ) \frac{\partial F}{\partial(\mathrm{\Delta}x)} = \left( \frac{1}{\sqrt{\mathrm{\Delta}y}} \right) ∂ ( Δ x ) ∂ F ​ = ( Δ y ​ 1 ​ ) and ∂ F ∂ ( Δ y ) = ( Δ x ) ∣ − 1 2 ( Δ y ) − 3 2 ∣ \frac{\partial F}{\partial(\mathrm{\Delta}y)} = (\mathrm{\Delta}x) \left| - \frac{1}{2}{(\mathrm{\Delta}y)}^{- \frac{3}{2}} \right| ∂ ( Δ y ) ∂ F ​ = ( Δ x ) ∣ ∣ ​ − 2 1 ​ ( Δ y ) − 2 3 ​ ∣ ∣ ​ , we can write:

δ F ( Δ x , Δ y ) = ( 1 Δ y ) ( δ Δ x ) 2 + 1 4 ( Δ x ) 2 ( Δ y ) − 3 ( δ Δ y ) 2   \delta F(\mathrm{\Delta}x,\mathrm{\Delta}y) = \sqrt{{(\frac{1}{\mathrm{\Delta}y}) (\delta\mathrm{\Delta}x)}^{2} + \frac{1}{4}{(\mathrm{\Delta}x)}^{2} (\mathrm{\Delta}y)^{- 3} {(\delta\mathrm{\Delta}y)}^{2}\ } δ F ( Δ x , Δ y ) = ( Δ y 1 ​ ) ( δ Δ x ) 2 + 4 1 ​ ( Δ x ) 2 ( Δ y ) − 3 ( δ Δ y ) 2   ​ (A2.4)

δ v o =   g 2 ( 1 Δ y ) ( δ Δ x ) 2 + 1 4 ( Δ x ) 2 ( Δ y ) − 3 ( δ Δ y ) 2   \delta v_{o} = \ \sqrt{\frac{g}{2}}\sqrt{{(\frac{1}{\mathrm{\Delta}y}) (\delta\mathrm{\Delta}x)}^{2} + \frac{1}{4}{(\mathrm{\Delta}x)}^{2} (\mathrm{\Delta}y)^{- 3} {(\delta\mathrm{\Delta}y)}^{2}\ } δ v o ​ =   2 g ​ ​ ( Δ y 1 ​ ) ( δ Δ x ) 2 + 4 1 ​ ( Δ x ) 2 ( Δ y ) − 3 ( δ Δ y ) 2   ​ (A2.5)

Now, we show how v o x v_{ox} v o x ​ and v o x v_{ox} v o x ​ are determined at different launch angles. Let v o x = v o c o s θ = G ( v o , θ ) v_{ox} = v_{o}{cos}\theta = G\left( v_{o},\theta \right) v o x ​ = v o ​ cos θ = G ( v o ​ , θ ) and v o y = v o s i n θ = H ( v o , θ ) v_{oy} = v_{o}{sin}\theta = H\left( v_{o},\theta \right) v oy ​ = v o ​ s in θ = H ( v o ​ , θ ) . Then, applying equation A1.9, we find that:

δ ( v o x ) = ( ∂ G ∂ v o ) 2 ( δ v o ) 2 + ( ∂ G ∂ θ ) 2 ( δ θ ) 2 = ( cos ⁡ θ ) 2 ( δ v o ) 2 + ( − v o sin ⁡ θ ) 2 ( δ θ ) 2 \delta\left( v_{ox} \right)=\sqrt{{(\frac{\partial G}{{\partial v}_{o}})}^{2} {(\delta v_{o})}^{2} + \left( \frac{\partial G}{\partial\theta} \right)^{2} {(\delta\theta)}^{2}} = \sqrt{{(\cos\theta)}^{2} {(\delta v_{o})}^{2} + \left( {- v}_{o} \sin\theta \right)^{2} {(\delta\theta)}^{2}} δ ( v o x ​ ) = ( ∂ v o ​ ∂ G ​ ) 2 ( δ v o ​ ) 2 + ( ∂ θ ∂ G ​ ) 2 ( δ θ ) 2 ​ = ( cos θ ) 2 ( δ v o ​ ) 2 + ( − v o ​ sin θ ) 2 ( δ θ ) 2 ​

= ∣ v o c o s θ ∣ ( δ ( v o ) v o ) 2 + ( ( s i n θ ) δ θ ) cos ⁡ θ ) 2 = \left| v_{o}{cos}\theta \right|\sqrt{\left( \frac{\delta\left( v_{o} \right)}{v_{o}} \right)^{2} + \left( \frac{{(sin}{\theta)} \delta\theta)}{\cos\theta} \right)^{2}} = ∣ v o ​ cos θ ∣ ( v o ​ δ ( v o ​ ) ​ ) 2 + ( c o s θ ( s in θ ) δ θ ) ​ ) 2 ​ (A2.6)

Similarly, applying equation (A2.2) to H ( v o , θ ) H\left( v_{o},\theta \right) H ( v o ​ , θ ) , we find,

δ ( v o y ) = ( ∂ H ∂ v o ) 2 ( δ v o ) 2 + ( ∂ H ∂ θ ) 2 ( δ θ ) 2 = ( sin ⁡ θ ) 2 ( δ v o ) 2 + ( v o cos ⁡ θ ) 2 ( δ θ ) 2 \delta\left( v_{oy} \right) = \sqrt{{(\frac{\partial H}{{\partial v}_{o}})}^{2} {(\delta v_{o})}^{2} + \left( \frac{\partial H}{\partial\theta} \right)^{2} {(\delta\theta)}^{2}} = \sqrt{{(\sin\theta)}^{2} {(\delta v_{o})}^{2} + \left( v_{o} \cos\theta \right)^{2} {(\delta\theta)}^{2}} δ ( v oy ​ ) = ( ∂ v o ​ ∂ H ​ ) 2 ( δ v o ​ ) 2 + ( ∂ θ ∂ H ​ ) 2 ( δ θ ) 2 ​ = ( sin θ ) 2 ( δ v o ​ ) 2 + ( v o ​ cos θ ) 2 ( δ θ ) 2 ​

= ∣ v o s i n θ ∣ ( δ ( v o ) v o ) 2 + ( ( c o s   θ ) δ θ sin ⁡ θ ) 2 = \left| v_{o}{sin}\theta \right|\sqrt{\left( \frac{\delta\left( v_{o} \right)}{v_{o}} \right)^{2} + \left( \frac{({cos\ \theta)}{\delta\theta}}{\sin\theta} \right)^{2}} = ∣ v o ​ s in θ ∣ ( v o ​ δ ( v o ​ ) ​ ) 2 + ( s i n θ ( cos   θ ) δ θ ​ ) 2 ​ (A2.7)

Next, we derive the equations necessary to determine the time of flight and its associated error. Using equation (6), the time of flight for the projectile launched at an angle ( θ ) \theta) θ ) from a reference height ( Δ y =   − h {\Delta}y = \ - h Δ y =   − h ) with coordinates (0, 0) is calculated from: − h = ( v o sin ⁡ θ ) t − 1 2 g t c a l 2 -h = (v_{o}\sin{\theta)} t - \frac{1}{2}g{t_{cal}}^{2} − h = ( v o ​ sin θ ) t − 2 1 ​ g t c a l ​ 2 as:

To find δ ( t c a l ) \delta\left( t_{cal} \right) δ ( t c a l ​ ) , we write equation (A2.8) as:

t c a l = 1 g [ δ ( v o y ) + U ( v o , θ ) ] t_{cal} = \frac{1}{g} \left\lbrack \delta\left( v_{oy} \right) + U\left( v_{o},\theta \right) \right\rbrack t c a l ​ = g 1 ​ [ δ ( v oy ​ ) + U ( v o ​ , θ ) ] (A2.9)

where U ( v o , θ ) = ( v o sin ⁡ θ ) 2 + K U\left( v_{o},\theta \right) = \sqrt{\left( v_{o} \sin\theta \right)^{2} + K} U ( v o ​ , θ ) = ( v o ​ sin θ ) 2 + K ​ and K = 2 g h . K = 2gh. K = 2 g h . Applying equation (A1.10) to equation A2.9,

δ ( t c a l ) = 1 g ( δ ( v o y ) ) 2 + ( δ U ( v o , θ ) ) 2 \delta\left( t_{cal} \right) = \frac{1}{g}\sqrt{{(\delta\left( v_{oy} \right))}^{2} + {(\delta U\left( v_{o},\theta \right))}^{2}} δ ( t c a l ​ ) = g 1 ​ ( δ ( v oy ​ ) ) 2 + ( δ U ( v o ​ , θ ) ) 2 ​ (A2.10)

Applying equation A1.9 to U ( v o , θ ) U\left( v_{o},\theta \right) U ( v o ​ , θ )

δ ( U ( v o , θ ) = ( ( v o sin ⁡ θ ) ( s i n θ ) δ v o ( v o sin ⁡ θ ) 2 + K ) 2 + ( ( v o 2 sin ⁡ θ ) ( cos ⁡ θ ) δ θ ( v o s i n θ ) 2 + K ) 2 \delta(U\left( v_{o},\theta \right) = \sqrt{{(\frac{(v_{o} \sin{\theta) {(sin}{\theta)} \delta v_{o}}}{\sqrt{{(v_{o} \sin{\theta)}}^{2} + K}})}^{2} + {(\frac{{{(v}_{o}}^{2} \sin{\theta)} \left( \cos\theta \right) \delta\theta}{\sqrt{{(v_{o} sin{\theta)}}^{2} + K}})}^{2}} δ ( U ( v o ​ , θ ) = ( ( v o ​ s i n θ ) 2 + K ​ ( v o ​ s i n θ ) ( s in θ ) δ v o ​ ​ ) 2 + ( ( v o ​ s in θ ) 2 + K ​ ( v o ​ 2 s i n θ ) ( c o s θ ) δ θ ​ ) 2 ​ (A2.11)

Substituting equations (A2.7) and (A2.11) in equation (A2.10),

δ t c a l = 1 g ( v o s i n θ ) 2 ( ( δ ( v o ) v o ) 2 + ( ( c o s   θ ) δ θ sin ⁡ θ ) 2 ) + ( ( v o sin ⁡ θ ) ( s i n θ ) δ v o ) 2 ( v o sin ⁡ θ ) 2 + K + ( ( v o 2 sin ⁡ θ ) cos ⁡ θ ) δ θ ) 2 ( v o sin ⁡ θ ) 2 + K {\delta t}_{cal} = \frac{1}{g}\sqrt{\left( v_{o}{sin}\theta \right)^{2} \left( \left( \frac{\delta\left( v_{o} \right)}{v_{o}} \right)^{2} + \left( \frac{({cos\ \theta)}{\delta\theta}}{\sin\theta} \right)^{2} \right) + \frac{{((v_{o} \sin{\theta) {(sin}{\theta)} \delta v_{o})}}^{2}}{{(v_{o} \sin{\theta)}}^{2} + K} + \frac{{({{(v}_{o}}^{2} \sin{\theta)} \cos\theta) \delta\theta)}^{2}}{{(v_{o} \sin{\theta)}}^{2} + K}} δ t c a l ​ = g 1 ​ ( v o ​ s in θ ) 2 ( ( v o ​ δ ( v o ​ ) ​ ) 2 + ( s i n θ ( cos   θ ) δ θ ​ ) 2 ) + ( v o ​ s i n θ ) 2 + K (( v o ​ s i n θ ) ( s in θ ) δ v o ​ ) 2 ​ + ( v o ​ s i n θ ) 2 + K ( ( v o ​ 2 s i n θ ) c o s θ ) δ θ ) 2 ​ ​

δ t c a l = 1 g [ v o sin ⁡ θ ( δ v o v o ) 2 + ( cot ⁡ θ δ θ ) 2 + ( ( s i n θ ) δ v o ) 2 ( v o sin ⁡ θ ) 2 + K + ( v o ( cos ⁡ θ ) δ θ ) 2 ( v o sin ⁡ θ ) 2 + K ] {\delta t}_{cal} = \frac{1}{g}\left\lbrack v_{o} \sin\theta\sqrt{\left( \frac{\delta v_{o}}{v_{o}} \right)^{2} + {(\cot{\theta \delta\theta)}}^{2} + \frac{{{((sin}{\theta)} \delta v_{o})}^{2}}{{(v_{o} \sin{\theta)}}^{2} + K} + \frac{\left( v_{o}\left( \cos\theta \right) \delta\theta \right)^{2}}{{(v_{o} \sin{\theta)}}^{2} + K}} \right\rbrack δ t c a l ​ = g 1 ​ [ v o ​ sin θ ( v o ​ δ v o ​ ​ ) 2 + ( cot θ δ θ ) 2 + ( v o ​ s i n θ ) 2 + K (( s in θ ) δ v o ​ ) 2 ​ + ( v o ​ s i n θ ) 2 + K ( v o ​ ( c o s θ ) δ θ ) 2 ​ ​ ]

δ t c a l = 1 g { v o sin ⁡ θ ( δ v o v o ) 2 + ( cot ⁡ θ δ θ ) 2 + ( ( s i n θ ) δ v o ) 2 ( v o sin ⁡ θ ) 2 + 2 g h + ( v o ( cos ⁡ θ ) δ θ ) 2 ( v o s i n θ ) 2 + 2 g h } {\delta t}_{cal} = \frac{1}{g}\left\{ v_{o} \sin\theta\sqrt{\left( \frac{\delta v_{o}}{v_{o}} \right)^{2} + {(\cot{\theta \delta\theta)}}^{2} + \frac{{{((sin}{\theta)} \delta v_{o})}^{2}}{{(v_{o} \sin{\theta)}}^{2} + 2gh} + \frac{\left( v_{o}\left( \cos\theta \right) \delta\theta \right)^{2}}{{(v_{o} sin{\theta)}}^{2} + 2gh}} \right\} δ t c a l ​ = g 1 ​ { v o ​ sin θ ( v o ​ δ v o ​ ​ ) 2 + ( cot θ δ θ ) 2 + ( v o ​ s i n θ ) 2 + 2 g h (( s in θ ) δ v o ​ ) 2 ​ + ( v o ​ s in θ ) 2 + 2 g h ( v o ​ ( c o s θ ) δ θ ) 2 ​ ​ } (A2.12)

Finally, the derivation of the error in ∆x cal is derived starting from   Δ x c a l = v o x t c a l \ {\mathrm{\Delta}x}_{cal} = v_{ox} t_{cal}   Δ x c a l ​ = v o x ​ t c a l ​ , to find

δ ( Δ x c a l c u l a t e d ) = ∣ Δ x c a l c u l a t e d ∣ ( δ v 0 x v o x ) 2 + ( δ t c a l t c a l ) 2 \delta\left( {\mathrm{\Delta}x}_{calculated} \right) = \left| {\mathrm{\Delta}x}_{calculated} \right| \sqrt{\left( \frac{\delta v_{0x}}{v_{ox}} \right)^{2}{+ \left( \frac{\delta t_{cal}}{t_{cal}} \right)}^{2}} δ ( Δ x c a l c u l a t e d ​ ) = ∣ Δ x c a l c u l a t e d ​ ∣ ( v o x ​ δ v 0 x ​ ​ ) 2 + ( t c a l ​ δ t c a l ​ ​ ) 2 ​ (A2.13)

Acknowledgements

The authors want to acknowledge and thank the School of Science and Technology at Georgia Gwinnett College for providing funds for the purchase of materials to support this activity.

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