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

Answer:

F = 15771.6 N

Explanation:

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

$v_i = \sqrt{2gh_1}$

$v_i = \sqrt{2(9.81)(32.2)}$

$v_i = 25.13 m/s$

now for final speed of rebound we have

$v_f = \sqrt{2gh_2}$

$v_f = \sqrt{2(9.81)(22.1)}$

$v_f = 20.82 m/s$

now the average force is given as

$F = \frac{mv_f - mv_i}{\Delta t}$

$F = \frac{0.556(20.82 + 25.13)}{1.62 \times 10^{-3}}$

$F = 15771.6 N$