A tennis ball is dropped from 1.16 m above the ground. It rebounds to a height of 0.866 m.
With what velocity does it hit the ground?
The acceleration of gravity is 9.8 m/s

Vf = 4.77 m/s

Explanation:

During the downward motion we can easily find the final velocity or the velocity with which the ball hits the ground, by using third equation of motion. The third equation of motion is given as follows:

2gh = Vf² - Vi²

where,

g = acceleration due to gravity = 9.8 m/s²

h = height = 1.16 m

Vf = Final Velocity of Ball = ?

Vi = Initial Velocity of Ball = 0 m/s (Since, ball was initially at rest)

Therefore, using these values in the equation, we get:

(2)(9.8 m/s²)(1.16 m) = Vf² - (0 m/s)²

Vf = √(22.736 m²/s²)

Vf = 4.77 m/s

he . if i may ask what i your question exactly?

explanation:

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.