Question

After falling from rest from a height of 27 m, a 0.46 kg ball rebounds upward, reaching a height of 17 m. If the contact between ball and ground lasted 2.1 ms, what average force was exerted on the ball?

Answer 1

Given m = 0.46 kg

In the first case h1 = 27 m

The velocity before hitting the ground is v1 = sqrt(2 * g * h1)

v1 = sqrt(2 * 9.8 * 27)

v1 = - 23.004 m/s (downward direction)

second case h2 = 17 m

The velocity after hitting the ground is

v2 = sqrt(2 * g * h2)

v2 = sqrt(2 * 9.8 * 17)

v2 = + 18.25 m/s (upward direction)

Given time t = 2.1 * 10^-3 s

from Impulse momentum theorem

Impulse = change in momentum

F * t = m * (v2 - v1)

F * 2.1 * 10^-3 = 0.46 * (18.25 - (-23.004))

F = 9036.59 N