Atennis ball is dropped from 1.43 m above the
ground. it rebounds to a height of 0.945 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
(this i already know -5.29 but i just don't know how to continue and do the rest of the parts)

part 2
with what velocity does it leave the ground?
answer in units of m/s.

part 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.

Part 2: 4.30 m/s

Part 3: 1503 m/s^2

Explanation:

First you already had -5.29 m/s

Part 2

With what velocity does it leave the ground ?

it is the same process as question 1, but in this case the height is 0.945 m

Part 3

This is a simple formula of :

Final speed: is the speed the ball leaves the ground = 4.30 m/s

Initial speed: is the speed the ball hits the ground = -5.29 m/s

4.30 - (-5.29) = 0.00638a

a = 1503 m/s^2

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.

did yall get answers for this yet