Question

A 0.75-kg ball falls vertically downward from a height of 55.0 m and rebounds upward. if the ball reaches a height of 30.0 m and the contact between the ball and ground lasts 2.50 ms, what is the average force exerted on the ball by the ground?

Answer 1

Impuls I is given by:
$I = \delta mv = |F|t$
where $\delta mv$ is the change in momentum, $|F|$ is the average force and t is the time.

Solve the equation for the force F:
$|F| = \frac{\delta mv}{t} = \frac{m(v_f - v_i)}{t}$

Energy should be conserved, so the velocities will be:
$\frac{1}{2}mv^2 = mgh \\ v = \sqrt{2gh}$

Combining both equations:
$|F| = \frac{m( \sqrt{2g(h_f + h_i)} )}{t}$
where $h_f = 30m, h_i = 55m, m = 0.75kg, t = 0.0025s$