Carpio_Erlinda - Projectile Motion Lab

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Treasure Valley Community College *

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GSCI 104

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Physics

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Apr 3, 2024

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GSCI 104 Name____ Erlinda Carpio _______ Projectile Motion PhET Lab https://phet.colorado.edu/en/simulation/projectile-motion Use only the Intro tab Part A: Horizontal Launch 1. Make sure you are using only a horizontal launch (angle = zero degrees) in this section. Fire the canon using the red launch button. Describe the path taken by the pumpkin. The pumpkin first travels horizontally and then curves downward. 2. Now change the mass of the object fired. Try several different objects. You may erase the path of the projectile using the eraser button. What happens to the path of the projectile when you change the mass of the projectile? The path of the different masses of each of the objects remains the same; travel horizontally and then curves downward. 3. Use the blue tool to measure the time, range and height of several points along the path. (Place the crosshairs onto a point to measure.) a. What is the time lapse between any two adjacent points? I found the time lapse of two points one point 0.6s and 0.7s, therefore the time lapsed would be 0.1s. b. Continue to measure successive points as you move away from the canon. i. What pattern do you notice in the range between successive points? As you progress from each point away from the cannon you will notice that it increases by 1.5m. ii. What pattern do you notice in the height between successive points? The height decreases. 4. Now change the initial speed of the projectile. Try several different speeds, and examine the paths using the blue tool like you did for #3. a. What is the relationship between the initial speed and the range? (“As the initial speed increases, the range….”) As I adjusted the speed, I found that the range would increase by 2m. c. What is the relationship between the initial speed and the time in flight? As I adjusted the speed, I noticed that the time would increase by 0.1s between each point. c. Compare the paths of flights with several different initial speeds. There is an open circle at the spot representing a flight time of exactly 1.0 second. What do you notice about this open circle in all of the different flight paths? The range is increasing by 1m for each increase 1m/s in initial speed.

5. Clear the screen by clicking on the yellow reset button. Now check the Air Resistance box. Compare the paths of a canon shot with and without Air Resistance. What happens to the a. range, b. height, and c. flight time when Air Resistance is acting? With air resistance at 0.1s the range is at 3m with a total range of 21.35m and a height of 9.95m at 0.1s. Without air resistance at 0.1s the range is at 1.5 m, with a total range of 21.42m and a height of 9.95m at 0.1s as well. The flight time remained constant at about 1.43s. d. Compare different objects. Does Air Resistance have the same influence on each projectile? Explain. Yes, each object changed with the addition of air resistance varied based on its drag coefficient. 6. Reset the simulation and check the “slow” button to slow the simulation down. (You may also click on the pause button to stop the projectile mid-flight.) Also, check the Velocity Vectors boxes. Fire the canon and pause the simulation several times along the projectile’s path. As the projectile moves farther from the canon, what pattern do you notice in the a. horizontal vector The vector remains the same during the flight, remaining parallel to the ground. b. vertical vector The vector expands in distance from 0 at the firing of the canon to 2.5m at 0.5s of flight, and to 5.0 m of height after the one second of flight. With that said, it remains perpendicular to the ground through the flight. c. Total vector? The total vector gradually changes its direction from a 90-degree angle from the vertical vector to eventually a 45-degree angle from the vertical vector throughout the flight. 7. Now repeat using the Acceleration Vectors boxes. What pattern do you notice? The horizontal vector remains about .75 meters throughout the flight and remains parallel to the ground in the opposite direction of the flight path. The vertical vector remains perpendicular to the ground but shortens in length from 5.0 m at the start of the flight to 4.76m at .5 seconds of flight and 4.3m 1.0 seconds of flight. The total vector remains the same with the angle of the vector towards the ground and slightly in the opposite direction. Part B: Angular Launch – use the same initial speed for each step below 8. Reset the simulation and tilt the canon to an angle of 35 o . Describe the path of the projectile when you fire the canon. The path of flight is longer. It takes 2.55 seconds to reach the ground and the range increases to 31.37 m. The height of the path has increased as well to 13.77m. 9. Keeping the angle at 35 o , move the canon stand to different heights. In words, compare and contrast the projectile paths at lower and greater starting heights. The apex of the trajectory consistently reduces and is directly related to the cannon’s elevation from the ground. This implies that a reduction in height by 2 results in a corresponding decrease in the peak height by two. As the cannon’s position is lowered, the flight duration decreases incrementally, with each adjustment leading to varying reductions that intensify progressively. Similarly, the distance covered (range) diminishes with each lowering of the cannon, and the difference in range becomes progressively smaller with each modification. 10. Now use the blue tool to measure corresponding points (such as the 5 th and 10 th points) on several different paths. What happens to the a. range, b. height, and c. flight time of corresponding points as you increase the height of the canon stand?

The flight time and the range remain constant at different points during the flight, however, the height is proportional to the increase in height of the cannon. 11. What happens to the total (final point in each path) a. range, b. height, and c. flight time as you increase the height of the canon stand? The total flight range and time increase as you raise the height of the canon stand and receives the biggest increase in time when it is changed from 2m to 4m and the biggest increase in range from 0m to 2m. 12. Lower the canon to a height of zero. How is the flight from this height different than the flight from any other height? (Note, in this simulation, the landing spot is always at a height of zero.) The peak height of the flight path indicated by the yellow circle and is lower than any other height of the canon and its flight path also lowers proportionally. 13. With the canon still at a height of zero, change the angle of the canon launch. Note the maximum times, heights, and ranges for several different launches. a. Which angle gives the maximum flight time? 90 degrees with 3.12s of flight time b. Which gives the maximum range? 45 degrees with 23.9m 14. a. With the canon still at a height of zero, compare and contrast the paths for projectiles shot at angles of 35 o and 55 o . What is similar and what is different? The flight times were different with 2.51 s for 55 degrees, and 1.75 s for 35 degrees. The peak height for 55 degrees was 5.73 m and was much higher than the 35-degree angle which was 3.77 m. The range on both 35 and 55 degrees were the same at 21.6m. b. Now compare and contrast the paths for projectiles shot at angles of 25 o and 65 o . What do you notice? At a degree angle of 25 the height was significantly lower where the highest reached 2.04m but with a range of 17.57m. At the degree angle of 65 the highest reached was 9.42m but ultimately reached the same range of 17.57m. The overall flight of the cannon of the 25 degree angle was 1.29s and the 65 degree angle was 2.77s. c. Suggest another pair of angles that would behave similarly. Try it out. Was your prediction correct? 30 degrees and 60 degrees Yes, both angles had the same range of 19.86 m. The peak height of the flight was very different with 8.6 m at 60 degrees, and 2.87 m at 30 degrees. The flight time was longer at 60 degrees with 2.65 s and 1.53 s at 30 degrees. Part C: Hit the Bullseye Try different combinations of Initial Speed, Launch Angle, Starting Height, Air Resistance, and Object Launched. Come up with at least three distinct methods for hitting the bullseye (three stars). List the all of the parameters used for each of the successful launches below. 1)Initial speed: 20 m/s, launch angle: 30 degrees, Starting height: 0m, Launched: Pumpkin Prediction ~35m and hit with 3 stars. 20cos30=17.3205080756 m/s 20sin30= 10.0 m/s Flight time =2 seconds D= v t 17.3205 x 2 = 34.6

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2) Initial speed: 10 m/s, launch angle: 45 degrees, starting height: 2m, Launched: Pumpkin Prediction ~12m and hit with 3 stars. 10cos45=7.07106781186 m/s 20sin30= 7.07106781186 m/s Flight time =1.41421356237 s D= v(t)+h 7.07106781186 x 1.4142356237 +2 =11.9999

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