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If the ping pong ball launcher is angled at 45 degrees and the distance it travels is 1.2 m. What would be the height it reaches? What time does it touch the ground

If the ping pong ball launcher is angled at 45 degrees and the distance it travels is 1.2 m. What would be the height it reaches? What time does it touch the ground

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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If the ping pong ball launcher is angled at 45 degrees and the distance it travels is 1.2 m. What would be the height it reaches? What time does it touch the ground? Please use this equation. Vic and Vy1 show in an explanation below
These components are displaced with distinct velocities due to the acceleration they experience. The ball undergoes type 3 projectile motion since it travels under the influence of gravity, it ejects at an angle of 8 degrees above the ground, and has an elevation of 0.491 m when launched. To find the initial velocity of the ball, the velocity vector must be broken down into Vx and Vy1. In this case, Vx = V cos8 and Vyl = V Sin8. Given that the maximum distance is 4m, the time of flight would be equal to 4/ v Cos8 in the horizontal direction. The time interval would be substituted into height + Vylt + 1/2 at - given 10m, and a = -9.8m/s. After solving for V, the ball initially travels with a velocity of 8.71 m/s @ 8. The horizontal velocity is 8.63 m/s [8.71cos8] and the formula dinitial = dinitial = Om, height %3D since the range is 4m, the time of flight is 0.463 seconds. The maximum height of the launch be calculated given that vy2 = 0 m/s, vy1 = 8.71sin8 = 1.21 m/s, can %3D (vy22 – vy1?)/2a) + dy. dy After calculations, the maximum height of the projectile is 0.566m. Finally, the impact = 0.491 m and a = -9.8 m/s - using the formula hmax = velocity can be calculated by first finding the vy2 using vy2 = Vvyl² + 2ad given that 9.8 m/s and d =-0.491m. After calculations, vy2 = 3.33 m/s vyl =- 8.71sin8 =- 1.21m/s, a =- To find the impact velocity, the horizontal and vertical components of velocity must be added [V impact = Vvx? + vy2 1 to give Vimpact = 9.25 m/s @ 21 ah. Refer to Figure 4 for a diagram of the projectile motion.
Transcribed Image Text:These components are displaced with distinct velocities due to the acceleration they experience. The ball undergoes type 3 projectile motion since it travels under the influence of gravity, it ejects at an angle of 8 degrees above the ground, and has an elevation of 0.491 m when launched. To find the initial velocity of the ball, the velocity vector must be broken down into Vx and Vy1. In this case, Vx = V cos8 and Vyl = V Sin8. Given that the maximum distance is 4m, the time of flight would be equal to 4/ v Cos8 in the horizontal direction. The time interval would be substituted into height + Vylt + 1/2 at - given 10m, and a = -9.8m/s. After solving for V, the ball initially travels with a velocity of 8.71 m/s @ 8. The horizontal velocity is 8.63 m/s [8.71cos8] and the formula dinitial = dinitial = Om, height %3D since the range is 4m, the time of flight is 0.463 seconds. The maximum height of the launch be calculated given that vy2 = 0 m/s, vy1 = 8.71sin8 = 1.21 m/s, can %3D (vy22 – vy1?)/2a) + dy. dy After calculations, the maximum height of the projectile is 0.566m. Finally, the impact = 0.491 m and a = -9.8 m/s - using the formula hmax = velocity can be calculated by first finding the vy2 using vy2 = Vvyl² + 2ad given that 9.8 m/s and d =-0.491m. After calculations, vy2 = 3.33 m/s vyl =- 8.71sin8 =- 1.21m/s, a =- To find the impact velocity, the horizontal and vertical components of velocity must be added [V impact = Vvx? + vy2 1 to give Vimpact = 9.25 m/s @ 21 ah. Refer to Figure 4 for a diagram of the projectile motion.
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