Lab 4 - projectile motion

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University of Ottawa *

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2315

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Psychology

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Dec 6, 2023

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APA 2315 LAB 4 - PROJECTILE MOTION Jeremiah Zephir 300130890 ( 70 marks ) OBJECTIVES: 1. To learn how to collect projectile motion data and calculate associated variables; 2. To evaluate the effect of release conditions on performance; 3. To apply knowledge of projectile motion to inform coaching decisions EQUIPMENT REQUIRED: Pencil, calculator, stop watches, 30 m tape, cones, football, balls INTRODUCTION: The term ballistic or projectile motion refers to the study of the behaviour of objects when they are thrown or projected into the air. Airborne objects are subjected to the laws of physics in a very specific manner. Over the course of this laboratory session you will explore a variety of experiments that will clarify how these concepts affect the trajectory of projectiles. A major influence on the trajectory of an airborne object is projection velocity . By increasing or decreasing projection velocity all other factors regulating the trajectory of the object will be greatly influenced. Along with projection velocity, projection angle has a large influence on the flight of any object. This occurs because velocity is a vector quantity . Vector quantities are defined by their magnitude (size), orientation (angle) and/or direction . In order to reach a given target, many different combinations of magnitudes and orientations are possible. Other factors, such as relative projection height , horizontal distance, vertical distance, accuracy, time limitations, and tactical issues also affect how we regulate ballistic motions during sporting activities. Many sports impose constraints on how the optimal velocity vector can be constructed for best performance. For example, in American and Canadian football, execution time for quarterbacks to successfully complete passes is very limited and therefore the options for the proper velocity vector are limited. At the other end of the spectrum, in golf, when obstacles are found between the player and the target, the type of trajectory that the ball needs to follow for optimal performance is severely limited by the environment. PART A – THROWING A FOOTBALL INSTRUCTIONS: In this laboratory, you will compute the projection angle of the velocity vector that allows for the successful completion of a football pass covering the length of half a volleyball court (9 m). To make these computations, you will calculate the average time needed to successfully complete 10 passes. Working in groups of 4 students, have one student throw a football 10 times to another student (same thrower and catcher each time), while the third student times the flight of the ball using a stopwatch, and a fourth participant records the times in the table below. Once 10 perfect passes have been achieved, follow the

APA 2315 – Introduction to the Biomechanics of Human Movement University of Ottawa 2 directions below to complete all pertinent calculations. To ensure the validity of your calculations, the ball must be thrown and caught at the same height. Time taken to cover horizontal distance of 9 meters: ( 1 mark ) Trial # Horizontal Distance (m) Flight Time (s) 1 9 0.83 2 9 0.91 3 9 1.06 4 9 0.90 5 9 1.06 6 9 1.08 7 9 1.03 8 9 0.95 9 9 0.96 10 9 0.85 Average 9 0.963 Use the information in the table above to calculate the following variables: 1. Horizontal velocity (v x ) ( 1 mark ) 2. Time taken for the ball to reach maximum height (t f ) ( 1 mark ) 3. Vertical velocity at take-off (= - vertical velocity at landing) (v y ) ( 1 mark ) 4. Maximum ball height (h) ( 1 mark ) 5. Projection angle ( ° ) ( 1 mark ) 6. Resultant velocity (v) ( 1 mark ) Calculation 1: Horizontal Velocity You can compute the horizontal velocity using this equation: 𝒅 ? = 𝒗 ? × 𝒕 where d h is the horizontal distance (in this case 9 m), v h is the horizontal velocity component (in m/s) and t is the average flight time (in s).

APA 2315 – Introduction to the Biomechanics of Human Movement University of Ottawa 3 v h = _________________ **Keep in mind that this calculation is only valid if ball was released and caught at the same height. Calculation 2: Time taken for ball to reach maximum height You can also use your data to compute the vertical component of the velocity vector. To do this, you must first divide the average flight time in half to determine how long it took your ball to reach its maximum height. You can do this because when the projectile is released and caught at the same height, the parabolic flight is symmetrical, and the time it takes for the ball to reach its maximum height is the same amount of time taken for the ball to return to its initial height. t 1/2 = _________________ Calculation 3: Vertical Velocity at take-off (= - vertical velocity at landing) A key characteristic of parabolic flight is that null (i.e. zero) vertical velocity is reached at maximum height. Further, the magnitude of the vertical component is the same at both ends of the flight path, however the directions of these components are opposite (at one end the ball is rising, at the other end i t’s falling). These characteristics of parabolic flight enable you to determine the vertical component of the velocity vector when the ball is released and caught, using the following equation: 𝒗 ? = 𝒗 ? + ?𝒕 where v f is the final velocity (in m/s), v i is the initial velocity (in m/s), g is gravitational acceleration (constant at -9.81 m/s 2 ) and t is the flight time (in s). v i = _________________ Calculation 4: Maximum ball height Another variable which can easily be computed from that data you have is the maximum height reached by the ball. To do so, you will need to use the following equation: ℎ = 𝑣 ? ? +1 2 ?? 2 where h is the maximum height reached by the ball (in m), v i is the initial velocity of the ball (in m/s), t is the time duration of the motion (in s), g is gravitational acceleration (in m/s 2 ).

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APA 2315 – Introduction to the Biomechanics of Human Movement University of Ottawa 4 h = _________________ **Food for thought: In this specific case, is h measured as a distance from the ground or from release height? ____________________________________________________ Calculation 5: Projection angle Using basic trigonometry, you can now compute the average projection angle that was used to successfully complete this experiment. In order to do so, basic assumptions supported by the characteristics of parabolic flight need to be made. One such assumption is that the maximum ball height was reached at the mid-point of the horizontal flight distance (i.e. 4.5 m). Using the velocity values that you computed earlier, you can use the tangent function to compute the projection angle. ???????? ?𝑎?𝜃 = 𝑎??𝑎???? where, in this example, opposite is the vertical velocity component (in m/s) and adjacent is the horizontal velocity component (in m/s). This relationship is based on the following characteristics of the right triangle: θ = _________________ Calculation 6: Resultant velocity Using the Pythagorean Theorem, you can compute the resultant velocity from the horizontal and vertical velocities. 𝑣 ? 2 + 𝑣 ? 2 = 𝑣 2 where v is the resultant velocity (hypotenuse), v y is vertical velocity in the y direction and v x the horizontal velocity in the x direction.

APA 2315 – Introduction to the Biomechanics of Human Movement University of Ottawa 5 v = ________________ QUESTIONS: 1. Why is it important that release and catch heights are the same? ( 1 mark ) Because the time of flight in projectile motion is dictated by the projectile's vertical velocity, it is crucial that the release and catch heights correspond. The projectile would have to travel a greater vertical distance if the release and catch heights were different, which would change the time of flight and cause it to be longer or shorter than anticipated. 2. How would changing only the release velocity affect flight time? ( 1 mark ) Increasing the release velocity while keeping other factors constant would generally result in a shorter flight time. This is because a higher initial velocity would cause the projectile to cover the horizontal distance more quickly, reducing the time it spends in the air. 3. How would changing only the maximum height affect flight time? ( 1 mark ) Changing only the maximum height, while keeping other factors constant, would not directly affect the flight time. The time of flight in projectile motion depends primarily on the initial velocity and the angle of projection. The maximum height reached by the projectile, however, is determined by the initial velocity and the gravitational acceleration. So, changing the maximum height would not impact the time of flight. 4. Give an example of a specific movement (include purpose of the movement) in a sport that would require a release angle of: ( 3 marks ) a. Greater than 45 : High jumpers in track and field use a release angle greater than 45 degrees when they perform the Fosbury Flop. The purpose of this movement is to clear the bar set at a specific height while minimizing the risk of knocking it off. A higher release angle allows the athlete to arch their body over the bar b. Less than 45 In baseball, when an outfielder throws the ball to home plate to prevent a runner from scoring, they typically use a release angle less than 45 degrees. This lower angle allows for a flatter trajectory, ensuring that the ball reaches the catcher quickly and accurately to prevent the runner from advancing.

APA 2315 – Introduction to the Biomechanics of Human Movement University of Ottawa 6 c. 45 A classic example of a release angle of 45 degrees is seen in basketball shooting. When a player shoots a jump shot from the free-throw line, they often aim for a 45-degree release angle. This angle maximizes the chance of the ball making a successful parabolic arc and entering the hoop. PART B – THROWING A BALL FOR MAXIMUM HEIGHT AND MAXIMUM DISTANCE INTRUCTIONS: For this portion of the lab, each participant will bounce a tennis ball off the ground twice with a different goal in mind for each. In the first trial, the participant will attempt to maximize the vertical height attained by the ball. In the second trial, the participant will attempt to maximize the horizontal distance (range) attained by the ball. In each case, the ball should be thrown with approximately the same amount of force. All calculations will be based on the trajectory between the two bounces. Do not throw the ball up in the air…bounce it off the gro und. For each condition, the flight time needs to be measured using a stopwatch, and the horizontal distance travelled by the airborne ball (i.e. the straight-line distance between the first & second bounces) needs to be measured using a measuring tape. If desired, prior to bouncing the ball, the tennis ball can be submerged in water so that the bounces leave visible water marks on the ground, making it easier to take accurate distance measurements. Upon completion of the throws, fill out the following chart, calculating the remaining unknown variables. Don’t forget to include your calculations in the appendix. ( 10 marks ) Condition Horizontal distance (d x ) Total flight time (t f ) Initial vertical velocity (V yi ) Horizontal velocity (v x ) Resultant initial velocity (v Ri ) Vertical distance (height) (d v ) Angle of release (θ) Maximized Height 4.18 m 2.09 s 10.25 2 10.01 5.35 78.47 Maximized Range 8.65 m 2.17 s 10.64 3.99 10.6 5.77 67.87 QUESTIONS: 1. How did you modify your throwing technique when throwing for maximum vertical height versus maximum horizontal distance? ( 2 marks ) Max vertical height: a factor that would influence both greatly would be initial velocity. Initial velocity would determine the initial power used, where power can result in greater distances. When it. Comes to height, the initial angle should be almost straight up at 90degrees. This allows the projectile to account for less horizontal distance and greater vertical Max horizontal range: again, the greater the initial speed the greater the distance can be covered. To achieve a max horizontal distance an ideal angle would be 45 degrees. This angle is optimal for

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APA 2315 – Introduction to the Biomechanics of Human Movement University of Ottawa 7 balancing between vertical and horizontal velocity, allowing the objest to travel the greatest horizontally all while stating in the air for as long as possibe. 2. After analyzing your data, what would you do to improve your performance? Why? ( 2 marks ) PART C - EFFECT OF RELEASE CONDITIONS (SPEED, ANGLE AND HEIGHT) ON PERFORMANCE For this portion of the lab, students will use an online projectile motion applet (available here: http://ngsir.netfirms.com/englishhtm/ThrowABall.htm ) to determine the effect of release speed, release angle and release height on the motion of an object. A control (or nominal) condition is listed first in the table below. Each of the subsequent conditions is modified to accommodate a 10% increase in speed, angle or height, respectively. Using the values listed below, fill in the blanks. ( 12 marks; 1 mark per velocity, 0.5 mark per flight time and horizontal distance ) Release Conditions Flight results Speed (m/s) Angle (°) Height (m) Final Vertical velocity (v v ) Final Horizontal velocity (v x ) Total flight time (t f ) Horizontal distance (d x ) Control 50.0 30.0 100.0 -50.85 43.3 7.74 335.15 +10% speed 55.0 30.0 100.0 -52.08 47.63 8.12 386.77 +10% angle 50.0 33.0 100.0 -51.95 41.93 8.08 338.82 +10% height 50.0 30.0 110.0 -52.76 43.3 7.93 343.48 Calculate the percent change in horizontal distance for each condition. ( 3 marks ) Percent difference = [(10% condition - control condition) / control condition)] * 100 QUESTIONS: 1. Which release variable was the horizontal distance most sensitive to? ( 1 mark ) Based on the equation used to solve for horizontal distance, are these the results you expected? Why? ( 2 marks ) Horizontal distance was most sensitive to changes in initial speed. The scenario with a 10% increase in speed resulted in the largest increase in horizontal distance with 15.4% difference.

APA 2315 – Introduction to the Biomechanics of Human Movement University of Ottawa 8 Regarding whether these results were expected based on the equation used to solve for horizontal distance, it aligns with our expectations. In projectile motion, horizontal distance is influenced by the horizontal velocity (which depends on speed and launch angle) and time of flight. An increase in speed directly increases the horizontal velocity, as a result leads to a greater horizontal distance. 2. The various release conditions led to different flight results. Are the changes to vertical speed, horizontal speed and total time what you expected based on the modifications made? Why? ( 9 marks ) 10% Increase in Speed: Vertical Speed: increasing speed typically results in an increase in both vertical and horizontal speed. Horizontal Speed: Increases due to the increase in speed. Total Time: May decrease slightly due to the increase in speed, which affects both horizontal and vertical motion. 10% Increase in Launch Angle: Vertical Speed: increasing the launch angle may lead to a slight increase in vertical speed. Horizontal Speed: May decrease slightly due to the increase in launch angle, as more of the initial velocity vector is directed upward. Total Time: May increase due to the higher launch angle, which results in a longer time of flight. 10% Increase in Initial Height: Vertical Speed: Increases due to the greater initial height. Horizontal Speed: Remains constant because the increase in height doesn't affect the horizontal speed. Total Time: Increases due to the higher initial height, leading to a longer time of flight. Increasing the speed tends to increase both vertical and horizontal speed and may lead to a shorter flight time. Increasing the launch angle may result in a longer flight time but could decrease horizontal speed slightly. Increasing the initial height tends to increase vertical speed and extend the flight time while leaving horizontal speed unaffected.

APA 2315 – Introduction to the Biomechanics of Human Movement University of Ottawa 9 PART D - CONSIDERATIONS FOR COACHING SHOT PUT ATHLETES BASED ON BODY TYPE For this portion of the lab, you will expand on knowledge gained in the previous sections to inform your decisions regarding coaching shot put athletes of various heights and muscular strengths. Of utmost importance for this part of the lab are: release height and release speed. Some helpful equations include: (⃑ℎ⃑⃑⃑ ? ⃑−⃑ℎ⃑⃑⃑ ? )= 𝑣⃑? ? sin 𝜃 + ?⃑? ? 2 ⃑?⃑⃑⃑ ? ⃑ = 𝑣⃑? ? cos 𝜃 𝑣⃑ 𝜃 ?????𝑎? where, θ optimal is the optimal release angle for achieving the greatest range v is the release velocity h i is the release height h f is the final height of the object t f is the flight time d x is the horizontal distance (range) g is the gravitational constant with a value of -9.81 m/s 2 Determine the optimal angle of projection (θ) and the corresponding horizontal displacement (d x ) under the following conditions: i. Athletes with different levels of muscular strength (represented by changes to speed) ( 6 marks ) Body type Speed (m/s) Height (m) Optimal θ (°) d x (m) Strong 12.00 2.10 49.8 12.4 Medium 8.00 2.10 57.1 3.85 Weak 4.00 2.10 38.54 2.68 ii. Athletes with different levels of muscular strength and varying body heights ( 6 marks ) Body type Speed (m/s) Height (m) Optimal θ (°) d x (m) Tall, Strong 12.00 2.10 49.8 12.4 Medium, Medium 8.00 1.900 57.13 4.2 Short, Weak 4.00 1.700 43.83 3.16

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APA 2315 – Introduction to the Biomechanics of Human Movement University of Ottawa 10 iii. Now let the weakest and shortest athlete use a lighter shot put (thus allowing for a greater release speed) ( 2 marks ) Body type Speed (m/s) Height (m) Optimal θ (°) d x (m) Short, Weak 8.00 1.700 55.31 4.5 QUESTIONS: 1. How does the athlete’s muscular strength (represented by the ability to release the shot put at a particular speed) affect optimal shot putting technique? ( 1 mark ) Greater muscular strength allows the athlete to generate more force, which, when applied correctly, can result in a higher initial speed for the shot put. This increased speed can contribute to a longer shot put distance. However, it's important to note that the optimal technique involves not only strength but also proper coordination and timing. 2. How does the athlete’s height (represented by release height) affect optimal shot putting technique? ( 1 mark. Leverage: Taller athletes have longer arms, which can provide greater leverage when pushing the shot put. This can result in a longer throwing arc and potentially more distance. Release Angle: A taller athlete may find it easier to release the shot put at an optimal angle for maximum distance. The release angle is influenced by the athlete's height and the trajectory they create during the throw. Gravity and Height: A taller athlete may have a slightly higher release point above the ground, which means they have a little more time to apply force to the shot put against the force of gravity. This can contribute to a longer throw.

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Lab 4 - projectile motion

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