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?
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