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
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 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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44bf5232-a6ed-4c3b-a009-8eef3cb0dcdc%2F8dce500c-b295-45ef-9cee-d4d6457984c2%2F5nswfwu_processed.jpeg&w=3840&q=75)
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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