The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²
The direction of the acceleration of the ball is downwards
The given parameters
initial velocity of the ball, u = 0
height above the ground, h = 2.2 m
time of motion of the ball, t = 96 ms = 0.096 s
The magnitude of the acceleration of the ball while coming to rest is calculated as;
let the downwards direction of the acceleration be positive

The direction of the acceleration of the ball is downwards
Learn more here: brainly.com/question/15407740
The elastic potential energy (Ep) is given by

Data:
Ep = ? (Joule)
k = 20 N/m
x (displacement) = 0.20 m
Solving:



