Question

After falling from rest at a height of 32.3 m, a 0.556 kg ball rebounds upward, reaching a height of 22.1 m. If the contact between ball and ground lasted 1.62 ms, what average force was exerted on the ball?

Answer 1

F = 15771.6 N

Step-by-step explanation:

Initial velocity of ball just before it will collide is given as

`v_i = √(2gh_1)`

`v_i = √(2(9.81)(32.2))`

`v_i = 25.13 m/s`

now for final speed of rebound we have

`v_f = √(2gh_2)`

`v_f = √(2(9.81)(22.1))`

`v_f = 20.82 m/s`

now the average force is given as

`F = (mv_f - mv_i)/(\Delta t)`

`F = (0.556(20.82 + 25.13))/(1.62 * 10^(-3))`

`F = 15771.6 N`