How do I find the velocity and acceleration

Question

A tennis ball is dropped from 1.43 m about the ground. It rebounds to a height of 0.945m. With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s^2. (Let down here negative.)

1.Answer in units of m/s.

2.With what velocity does it leave the ground? Answer in units of m/s

3. If the tennis ball were in contact with the ground for 0.00638 s, find the acceleration given to the tennis ball by the ground. Answer in units of m/s^2

Answer 1

The first two parts of the question - velocity when hits ground and velocity when bounces back can both be found from conservation of energy, assuming that air resistance is neglected.

Potential Energy: m g h

Kinetic energy: (1/2) m v²

For the downward direction:

E_i = - m g h + 0 (dropped from rest)

E_f = 0 + (1/2) m v²

By conservation of energy, E_i = E_f --> - m g h = (1/2) m v² cancel the mass from both sides of the equation and solve for v: v = √( 2 g h) = sqrt[2(9.8 m/s²)(1.43 m) ] = 5.29 m/s downward

The upward direction is identical, except E_i is the kinetic energy (starts out moving) and E_f is the potential energy (stops) and the height is 0.945 m. (v = 4.30 m/s upward)

The third part requires the use of impulse. Δp = I = F Δt, Δp is the change in momentum and Δt is the time the ball is in contact with the ground. F = ma (Newton's First Law) and momentum, p = mv

making the substitutions and rearranging: m a = Δp / Δt = (m Δv) / Δt --> a = Δv / Δt

Δv = v_f - v_i, where v_f is the velocity the instant the ball leaves the ground and v_i is the velocity the instant the ball hits the ground.

a = (4.30 m/s - (-5.29 m/s)) / (0.00638 s) = 1500 m/s²