020 (part 1 of 3) 10.0 points
A tennis ball is dropped from 1.97 m above the
ground. It rebounds to a height of 0.901 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s
2
. (Let
down be negative.)
Answer in units of m/s.
021 (part 2 of 3) 10.0 points
With what velocity does it leave the ground?
Answer in units of m/s.
022 (part 3 of 3) 10.0 points
If the tennis ball were in contact with the
ground for 0.0102 s, find the acceleration
given to the tennis ball by the ground.
To find the velocity with which the tennis ball hits the ground, we can use the equation for freefall motion:
v = sqrt(2gh)
where v is the final velocity, g is the acceleration due to gravity, and h is the height from which the ball is dropped.
In this case, the height h is given as 1.97 m and the acceleration due to gravity g is given as 9.8 m/s^2.
Plugging in these values, we get:
v = sqrt(2 * 9.8 * 1.97)
= sqrt(38.404)
≈ 6.2 m/s
Therefore, the velocity with which the tennis ball hits the ground is approximately 6.2 m/s.
To find the velocity with which the ball leaves the ground, we can use the principle of conservation of mechanical energy. The potential energy at the highest point will be equal to the kinetic energy just before hitting the ground.
Potential energy at highest point = Kinetic energy just before hitting the ground
mgh = (1/2)mv^2
where m is the mass of the tennis ball, g is the acceleration due to gravity, h is the height from which the ball is dropped, and v is the velocity just before hitting the ground.
Since the mass of the tennis ball cancels out, the mass is not required and we can simplify the equation as follows:
gh = (1/2)v^2
Now, we can find the velocity v using this equation. Plugging in the known values of height h as 0.901 m and acceleration due to gravity g as 9.8 m/s^2:
(9.8)(0.901) = (1/2)v^2
v^2 = 8.9849
v ≈ 2.997 m/s
Therefore, the velocity with which the tennis ball leaves the ground is approximately 2.997 m/s.
To find the acceleration given to the tennis ball by the ground, we can use the formula:
a = (v - u) / t
where a is the acceleration, v is the final velocity, u is the initial velocity, and t is the time taken.
In this case, the initial velocity u is 2.9978 m/s (which is the velocity with which the ball leaves the ground), and the time t is given as 0.0102 s.
Since the tennis ball starts from rest when it hits the ground, the initial velocity u is 0 m/s. Plugging in the values in the formula, we get:
a = (2.997 - 0) / 0.0102
≈ 294.76 m/s^2
Therefore, the acceleration given to the tennis ball by the ground is approximately 294.76 m/s^2.