Question

A tennis 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.

Answer 1

Answer:

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

$v =\sqrt{2gh}$

$v =\sqrt{2*9.8*0.945} \\v =4.30\ m/s$

Part 3

This is a simple formula of :

$v_{f}-v_{i}=at \\where \\v_{f}: final\ speed \\v_{i}: initial\ speed \\a: acceleration\\t: time$

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