Lab 04 - Projectile_updated

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PHYS 1: Fall 2023 Lab 04 KINEMATICS IN TWO DIMENSIONS Investigating 2D Motion: Objects under the Influence of Gravity Objective: This virtual lab activity is intended to enhance your physics understanding. It will help you make connections between predictions and conclusions, concepts and actions, equations, and practical activities. We also think that if you engage with this activity, it will be fun as well! This is an opportunity to learn a great deal. You should answer all questions as you follow the procedure in running the simulations. Click on the link below and then select the ‘ Lab ’ tab. https://phet.colorado.edu/sims/html/projectile-motion/latest/projectile-motion_en.html Use simulation controls from the bottom right controls. Click “Fire (Red Rectangle on the bottom left)” to launch the projectile or click “Erase” (next to “Fire”) to clear the projectile. You can pick different objects to shoot out of the canon by using the object selector from the top right. Fall 2023

PHYS 1: Fall 2023 You can manually adjust the settings of the projectile from the middle right projectile controls. Using projectile controls, you can set the angle, initial speed, mass, and diameter. If you wish to resemble real-world conditions, check the ‘Air resistance’ box. You can also add sound to the simulation by checking the sound box. Once you create a trajectory of a projectile, drag the blue rectangular box from the top right into the screen and place the crosshair of that rectangle along the trajectory to determine the Time, Range, and Height of the projectile at that instant. Range and height can also be verified using the “ Tape ” located on the top right. To move the tape measure, click (and hold) and drag it to the location of your choice. You can ‘ elongate ’ the tape by clicking and dragging it on the end of the tape. You can make a ‘game out of this simulation by trying to hit a target. In general, familiarize yourself with control features and displayed results. Introduction: The basic kinematics equations in one-dimensional motion are also used for two-dimensional motions. Since the two-dimensional motion is described using two components, x and y independently, the basic two-dimensional kinematics formula can be written as follows: When working with projectiles, we apply these kinematics equations with the following settings: ● An initial velocity, v 0, and initial (launch) angle θ o Horizontal component for the initial velocity is v 0 cos θ o Vertical component for the initial velocity is v 0 sin θ ● There is no acceleration in the horizontal direction: a x = 0 ● Gravitational acceleration is directed downwards: a y = -g The velocity at any point on the projectile results by applying the Pythagorean Theorem: 𝑣 = 𝑣 ? 2 + 𝑣 ? 2 The angle θ the velocity vector makes with the horizontal can be found using the following formula: = ??? −1 𝑣 ? 𝑣 ? ( ) When the kinematics equations are applied with the given specifications, the following useful equations can be derived for the case of projectiles fired from the ground . Range of the projectile: 𝑅 = 𝑣 0 2 𝑔 ?𝑖?(2θ) Fall 2023

PHYS 1: Fall 2023 Maximum height: ? ??? = 𝑣 0 2 2𝑔 ?𝑖?θ 2 Total time of flight: ? ????? = 2𝑣 0 ?𝑖? θ 𝑔 Projectile fired from a certain height: Figure 2 shows the trajectory of the projectile fired from height y 0 with an initial speed of v 0 at an angle of θ with the horizontal. The initial X- and Y- component of the velocity of the projectile is given by: ...... (3) 𝑣 0? = 𝑣 0 ???θ ....... (4) 𝑣 0? = 𝑣 0 ?𝑖?θ At any time “t”, the X- and Y- component of velocities are: ............... (5) 𝑣 ? (?) = 𝑣 0 ???θ ......... (6) 𝑣 ? ? ( ) = 𝑣 0? ?𝑖?θ − 𝑔? Where, is the acceleration due to gravity. 𝑔 = 9. 81?/? 2 If the projectile takes time “ t ” to fall to the ground, then the range (horizontal distance traveled) of the projectile is given by ..... (7) 𝑅 = 𝑣 0? × ? = 𝑣 0 ???θ×? Let us now consider just the Y - component of projectile motion. Then using kinematics equations, we can show that during this ‘free fall’, the vertical distance traveled by the projectile during time interval “t” is given by the equation, ...... (8) ? − ? 0 = 𝑣 0? ? − 1 2 𝑔? 2 At the instant when the projectile hits the ground, Equation (8) can be written as ...... (9) − 1 2 𝑔? 2 + 𝑣 0? ? + ? 0 = 0 This is equation a quadratic equation of the form: , which ?? 2 + ?? + ? = 0 we can solve to determine “ t ”. Once the time is calculated, the range, R , of the projectile can be determined using Equation (7). Fall 2023

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PHYS 1: Fall 2023 Procedure for the Experiment (Simulations): Section I: Projectile fired from the ground https://phet.colorado.edu/sims/html/projectile-motion/latest/projectile-motion_en.html Part I 1. Settings: “Reset/Erase” all previous settings. Maximize the screen size. Select ‘ pumpkin ’ from the top right. Set the projectile angle to 55 degrees and the initial speed to 18 m/ s. Keep the ‘launch-height’ at 0.0 m, mass and diameter remain in the default setting. Also ‘turn-off’ the air resistance. 2. “Launch” the pumpkin. 3. Drag the tape measure located on the top center of the screen to measure the ‘maximum-height’ and the range of the projectile. Record your measurements. Maximum height (y max ) = 11.08 m Range (x) = 30.50 m [3 + 3 = 6 Points] 4. Calculate the ‘ maximum height ’ and the range using the angle and initial speeds set at step 1 using formulas provided in the ‘Introduction’ section. You must show your work to get credit. [4 + 4 = 8 points] Calculated Maximum height (y max ) = 11.08 m ● = = 11.08 ? ??? = 𝑣 0 2 2𝑔 ?𝑖?θ 2 ? ??? = 18 2 2 (9.81) ?𝑖?(55) 2 Calculated Range (x) = 31.04 m ● = 𝑅 = 𝑣 0 2 𝑔 ?𝑖?(2θ) 𝑅 = 18 2 9.81 ?𝑖?(2(55)) = 31. 04 5. Is there a difference between your measurement and your calculated result? Compare the calculated results with the measurements. Explain with the reasoning if there are differences. [2 Points] ● There is a difference between what I calculated and my actual measurements. I believe that the differences are because the calculated results are based on idealized physics equations, assuming no air resistance and perfect conditions. In contrast, the measured simulation results are influenced by the accuracy and settings of the simulation itself. Small discrepancies between the idealized calculations and real-world simulations can occur due to factors such as simulation accuracy and the handling of physics within the software. Rounding in the calculations and measurements can lead to minor differences between the values. Fall 2023

PHYS 1: Fall 2023 Part II 6. Keep the settings in part I of step 1 except change the angle to 35 degrees. 7. Use the tape measure to find the maximum height and range. Record your measurement. Measured Maximum height (y max ) = 5.43 m Measured Range (x) = 30.90 m [3 + 3 = 6 Points] 8. Explain similarities and differences between the 55- and 35-degree settings. [2 Points] ● In the comparison between the two projectile motion settings, 55 degrees and 35 degrees, several key differences and a notable similarity emerge. Both settings share the same initial speed of 18 m/s, ensuring an identical starting velocity for the projectiles. However, the launch angles differ significantly, with the 55-degree setting featuring a steeper launch angle compared to the 35-degree setting. This variation in launch angle leads to distinct outcomes in the trajectory. In terms of maximum height, the 55-degree setting yields a higher calculated value of approximately 11.08 meters, while the 35-degree setting results in a measured maximum height of 5.43 meters. The 55-degree setting also boasts a slightly longer range, with a calculated range of about 31.04 meters, as opposed to the 30.50 meters measured in the 35-degree setting 9. Which setting has a longer time of flight? (Use the formula from ‘Introduction’). Show your calculations in order to earn full credit. [4 + 4 = 8 Points] ● The 55 degree angle has the longer flight time, as shown below in the calculations. 55 degrees has a flight time of 3.01 seconds while 35 degree angle has a flight time of 2.10 seconds. Total time (t) with 55-degree setting = 3.01s ● ? ????? = 2𝑣 0 ?𝑖? θ 𝑔 ? ????? = 2(18)?𝑖? 55 9..81 = 3. 01 Total time (t) with 35-degree setting = 2.10s ● = ? ????? = 2𝑣 0 ?𝑖? θ 𝑔 ? ????? = 2(18)?𝑖? 35 9..81 = 2. 10 Fall 2023

PHYS 1: Fall 2023 Part III 10.Clear the projectiles by clicking the “Erase” button 11. Go back to part I / step 1 settings and launch the pumpkin 12. Do not click the Erase button. 13. Now, check the air resistance box (drag coefficient automatically sets to 0.6) and launch the pumpkin. 14. Measure the maximum height and the range once air resistance is turned on. Measured Maximum height ( y max ) = 4.98 m Measured Range (x) = 26.28 [3 + 3 = 6 points] 15. Did you see the changes in height and range due to the air resistance? If so, how did the air resistance affect the height and range of the projectile? [2 Points] ● Yes, there are observable changes in the maximum height and range of the projectile when air resistance is turned on. With air resistance in effect, both the maximum height and range are reduced compared to when air resistance was turned off. The presence of air resistance has a lowering effect on the projectile's motion. As the projectile travels through the air, it experiences a resistive force due to air molecules colliding with it. This resistive force opposes the projectile's motion, causing it to lose energy and slowing it down. Thus, the projectile does not reach the same maximum height and covers a shorter horizontal distance before hitting the ground. 16. What effect do you see on the range and height if (keeping the drag coefficient constant)? a. Mass is increased as follows: [5 Points] Mass (kg) Max. Height (m) Range 5 4.98 13.68 10 5.19 14.53 15 5.27 14.84 20 5.31 15.00 25 5.33 15.10 Plot two graphs, one for Mass vs. Max. Height (y-axis) and another for Mass. Vs. Range (x-axis) in Excel and Paste below. Explain the variation (Is it linear? Quadratic?) ( Helpful video : https://www.youtube.com/watch?v=Vk8F99Ptq2E ) (All graphs should have axis labels and titles to get full credit.) [5 Points] ● There is a strong linear correlation between mass and range. Based on the r 2 values and the typical interpretation of such values, it appears that both relationships (Mass vs. Max. Height and Mass vs. Range) are linear within the given range of data. There is no curvature indicicating a quadratic. Fall 2023

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PHYS 1: Fall 2023 b. Diameter increased for fixed mass (Choose your mass M= … . kg) [5 Points] Diameter (m) Max. Height (m) Range 0.1 5.40 30.60 0.3 5.12 27.81 0.5 4.67 23.66 0.7 4.16 19.56 0.9 3.66 16.08 Fall 2023

PHYS 1: Fall 2023 Plot two graphs: One for Diameter vs. Max. Height (y-axis) and another for Diameter Vs. Range (y-axis) in Excel and Paste below. (Is it linear? Quadratic?) [5 Points] ● Based on this set of data, the relationship between diameter and max height is linear. This displayed linear relationship suggests that, within the range of data points that were collected, an increase in diameter corresponds to a proportional change in both maximum height and range, this is further proven by the high r^2 value of 0.989 and 0.996. Fall 2023

PHYS 1: Fall 2023 Section II: Projectile fired from a certain height Reset the simulation. Launch a ‘ Piano ’ at an angle of 30 degrees and complete the first row of the table. Now, click and drag the “+” sign located on the Canon to raise it by 1 m and complete the second row of the table. Show your calculation below for all Time of flights to get full credit. (Use equation (9) to calculate the time of Flight here!). [5 Points for Table and 10 points for showing calculation for ‘T’]. Launched from height (m) Max. Height Reached (m) Range (m) Time of Flight (T) 0 4.13 28.60 3.66 1 5.13 30.24 3.72 2 6.13 31.73 3.78 3 7.13 33.09 3.83 4 8.13 34.29 3.88 ● − 1 2 𝑔? 2 + 𝑣 0? ? + ? 0 = 0 ● ?? 2 + ?? + ? = 0 ● − 1 2 (9. 81) ? 2 + (18)? + 0 = 0 ○ − 4. 905 ? 2 + (18)? + 0 = 0 ■ t=3.66 s ● − 1 2 (9. 81) ? 2 + (18)? + 1 = 0 ○ − 4. 905 ? 2 + (18)? + 1 = 0 ■ t=3.72 s ● − 1 2 (9. 81) ? 2 + (18)? + 2 = 0 ○ − 4. 905 ? 2 + (18)? + 2 = 0 ■ 3.78 s ● − 1 2 (9. 81) ? 2 + (18)? + 3 = 0 ○ − 4. 905 ? 2 + (18)? + 3 = 0 ■ 3.83 s ● − 1 2 (9. 81) ? 2 + (18)? + 4 = 0 ○ − 4. 905 ? 2 + (18)? + 4 = 0 ■ 3.88 s Fall 2023

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PHYS 1: Fall 2023 Plot a graph of “Launched from height” vs. Time of Flight. Explain the variation. [5 Points] ● The data shows a clear positive linear correlation: when the initial height is greater, the time it takes for the projectile to reach the ground (Time of Flight) is also greater. The r 2 value close to 1 (0.998) indicates an extremely strong correlation between the initial height from which the projectile is launched and the time it takes for the projectile to reach the ground. The presence of air resistance can lead to variations in the relationship between initial height and time of flight, which may include factors such as the shape and mass of the projectile, its initial velocity, and the magnitude of the air resistance force. Fall 2023

PHYS 1: Fall 2023 Numerical Problem: (10 Points) Now let us combine the knowledge and skills we got from the simulation to solve this problem. Show your work to get credit. A player kicks a football at an angle of 37 0 with the horizontal and with an initial speed of 16 m/s. A second player standing at a distance of 33m from the first in the direction of the kick starts running to meet the ball at the instant it is kicked. Assuming the player is already running at a constant speed, how fast must he run in order to catch the ball before it hits the ground? ● 𝑅 = 𝑣 0 2 𝑔 ?𝑖?(2θ) ● 𝑇 = 2𝑣 0 ?𝑖? θ 𝑔 ● ? = 𝑇 = 2𝑣?𝑖?(θ) 𝑔 = 2 (16 ?/?)(?𝑖?(37)) 9.81 = ○ 1. 96? 𝑖? ?ℎ𝑒 ?𝑖? ● = 𝑅 = 𝑣 0? × ? = 𝑣 0 ???θ×? 16?/? (???(37) ? 1. 96? = ○ 25.05 m from initial position ● For him to catch the ball at his distance of 33 m, he would need to run at least 4.08 m/s ○ 33 m -25 m= = 4.08 m/s 8 1.96 Fall 2023

Lab 04 - Projectile_updated

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