After falling from rest from a height of 26 m, a 0.49-kg ball rebounds upward, reaching a height of 16 m. If the contact between ball and ground lasted 1.8 ms, what average force was exerted on the ball?

Answer 1

Divide the momentum change by the contact time.

Since momentum is a vector that changes You can compute the momentum going down and up, at impact, from the heights H1 and H2 reached by the ball.
Speed before impact = sqrt(2*g*H1) = 22.6 m/s
Speed after impact = sqrt(2*g*H2) = 17.7 m/s
Momentum change = (0.49)*(22.6+17.7) = 19.7 kg*m/s

Average force = 19.7/1.8*10^-3 = 10,970 N