Physics and Rebounds

Question

A ball of mass 0.745 kg is dropped from rest from a height of 3.75 m. It rebounds from the ground and reaches a height of 1.78 m, what impulse (specify magnitude and direction) was delivered to the ball by the ground? 

Answer 1

FΔt=mΔv is the relationship to use. The left side is the impulse and the right side is the change in momentum. In the above problem the ball will reach a velocity of (2gh)^1/2=(2*9.81*3.75)^1/2 = 8.58 m/s the instant before it strikes the ground. After the strike, the velocity is (2gh)^1/2 = (2*9.81*1.78)^1/2 = 5.91 m/s. The change in momentum as a result of striking the ground mΔv = .745(5.91-(-8.58))= 10.8 kg-m/s. (assume positive is up). Since Impulse = Change in Momentum the Impulse = 10.8 kg-m/s = 10.8 N-s.