Question

part 1 of 3

A tennis ball is dropped from 1.79 m above the ground. It rebounds to a height of 1.01 m. 

With what velocity does it hit the ground? The acceleration of gravity is 9.8 m/s². (Let down be negative.) 

Answer in units of m/s. 

part 2 of 3 

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

part 3 of 3 

If the tennis ball were in contact with the ground for 0.0126 s. find the acceleration given to the tennis ball by the ground. 

Answer in units of m/s².

Answer 1

a)

feom 3rd equation of motion

v^2 = 2 g h = 2 * 9.8* 1.79

v = 5.329 m/s

=====

b)

similarly

v ' ^2 = 2* 9.8* 1.01

v' = 4.449 m/a

=========

c)

a = ( v' - v) / t= ( 4.449 - 5.329) / 9.8

a = 0.089 m/s^2

=======

Comment before rate in case any doubt, will reply for sure.. goodluck