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If a ping pong launcher is angled at 45 degrees and goes a distance of 1.2 m. Please calculate the height and the time at which it touches your he ground. Please follow the format below

If a ping pong launcher is angled at 45 degrees and goes a distance of 1.2 m. Please calculate the height and the time at which it touches your he ground. Please follow the format below

Inquiry into Physics
Inquiry into Physics
8th Edition
ISBN:9781337515863
Author:Ostdiek
Publisher:Ostdiek
Chapter2: Newton's Laws
Section: Chapter Questions
Problem 2PIP: What important mathematical “tool” did Newton develop that made possible many of his discoveries in...
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If a ping pong launcher is angled at 45 degrees and goes a distance of 1.2 m. Please calculate the height and the time at which it touches your he ground. Please follow the format below
2 I 3 I ..4 . 5 Connections to Physics 7. The most substantial factor behind the COVID-Catapult's function is that there are several components in motion. For example, the arm can move by rotating, and the ping pong ball itself travels as a projectile during the majority of the launch. 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 = VSin8. 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 the formula dinitial = height + Vylt + 1/2 at given = -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 dinitial = 0m, height = 10m, and a since the range is 4m, the time of flight is 0.463 seconds. The maximum height of the launch can be calculated given that vy2 = 0 m/s, dy = 0.491 m and a = -9.8 m/s- using the formula hmax = ((vy2 - vyl)/2a) + dy. After calculations, the maximum height of the projectile is 0.566m. Finally, the impact vy1 = 8.71sin8 = 1.21 m/s, %3D velocity can be calculated by first finding the vy2 using vy2 = Vvy12 + 2ad given that vyl =-8.71sin8 =-1.21m/s, a =-9.8 m/s and d =-0.491m. After calculations, vy2 = 3.33 m/s . To find the impact velocity, the horizontal and vertical components of velocity must be %3D added [Vimpact = Vvx? +vy2 ] to give V impact = 9.25 m/s @ 21 ah. Refer to Figure 4 for a diagram of the projectile motion.
Transcribed Image Text:2 I 3 I ..4 . 5 Connections to Physics 7. The most substantial factor behind the COVID-Catapult's function is that there are several components in motion. For example, the arm can move by rotating, and the ping pong ball itself travels as a projectile during the majority of the launch. 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 = VSin8. 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 the formula dinitial = height + Vylt + 1/2 at given = -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 dinitial = 0m, height = 10m, and a since the range is 4m, the time of flight is 0.463 seconds. The maximum height of the launch can be calculated given that vy2 = 0 m/s, dy = 0.491 m and a = -9.8 m/s- using the formula hmax = ((vy2 - vyl)/2a) + dy. After calculations, the maximum height of the projectile is 0.566m. Finally, the impact vy1 = 8.71sin8 = 1.21 m/s, %3D velocity can be calculated by first finding the vy2 using vy2 = Vvy12 + 2ad given that vyl =-8.71sin8 =-1.21m/s, a =-9.8 m/s and d =-0.491m. After calculations, vy2 = 3.33 m/s . To find the impact velocity, the horizontal and vertical components of velocity must be %3D added [Vimpact = Vvx? +vy2 ] to give V impact = 9.25 m/s @ 21 ah. Refer to Figure 4 for a diagram of the projectile motion.
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