A tennis ball is dropped from 1.41m above the ground. It rebounds to a height of 0.932 m.The acceleration of gravity is 9.8 m/s2 .(a)With what velocity does it hit the ground?(Let down be negative.) Answer in units of m/s. (b)With what velocity does it leave the ground? Answer in units of m/s. (c)If the tennis ball were in contact with the ground for 0.00575 s, find the acceleration given to the tennis ball by the ground. Answer in units of m/s2.
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To solve these questions, we can use the equations of motion. Let's break down each part:
(a) With what velocity does the tennis ball hit the ground?
To find the velocity, we can use the equation of motion:
Vf^2 = Vi^2 + 2ad
Where:
- Vf is the final velocity when the ball hits the ground
- Vi is the initial velocity (which we need to find)
- a is the acceleration due to gravity (-9.8 m/s^2 in this case)
- d is the displacement (which is the distance between the initial height and ground)
Given that the initial height (drop height) is 1.41 m and using the downward direction as negative, the displacement becomes -1.41 m.
Plugging in all the values into the equation, we have:
0 = Vi^2 + 2(-9.8)(-1.41)
Simplifying, we get:
0 = Vi^2 + 27.3648
Rearranging the equation, we have:
Vi^2 = -27.3648
Vi = sqrt(-27.3648) = -5.230 m/s
Therefore, the velocity at which the tennis ball hits the ground is approximately -5.230 m/s (negative sign indicates downward direction).
(b) With what velocity does the tennis ball leave the ground?
When the ball rebounds, we can use the same equation of motion as before. However, in this case, the initial height will be given by the rebound height of 0.932 m.
Using the same equation of motion:
Vf^2 = Vi^2 + 2ad
We'll solve for Vi this time. The displacement 'd' is now positive since the ball is moving upwards, so d = 0.932 m.
Plugging in the values, we have:
0 = Vi^2 + 2(9.8)(0.932)
Simplifying:
0 = Vi^2 + 18.2392
Rearranging:
Vi^2 = -18.2392
Vi = sqrt(-18.2392) = -4.271 m/s
Therefore, the velocity at which the tennis ball leaves the ground is approximately -4.271 m/s (in the upward direction).
(c) If the tennis ball were in contact with the ground for 0.00575 s, find the acceleration given to the tennis ball by the ground.
To find the acceleration, we can use the equation:
a = (Vf - Vi) / t
Where:
- a is the acceleration
- Vf is the final velocity (which we need to find)
- Vi is the initial velocity (-4.271 m/s in this case)
- t is the time in contact with the ground (0.00575 s)
Plugging in the values, we have:
a = (Vf - (-4.271)) / 0.00575
Simplifying:
a = (Vf + 4.271) / 0.00575
Therefore, the acceleration given to the tennis ball by the ground depends on the final velocity (Vf). If you have the value of Vf, substitute it into the equation to find the acceleration.
(c)
Force = rate of change of momentum.
Calculate the momentum change, divide by the contact duration, you will get the acceleration, after the mass cancels out.
It will be a huge number over 1600 m/s/s.
See also following link for your reading:
http://www.vias.org/physics/example_2_5_13.html