a .140 kg baseball is dropped from rest from a height of 2.2m above the ground. it rebounds to a height of 1.6 m. what change in the balls momentum occurs when the ball hits the ground

Answer 1

The momentum change is
M [V1 -(-V2)] = M(V1 + V2)
For the two speeds, use
V1 = sqrt[g*2.2/(2M)] = 8.77 m/s
V2 = sqrt[g*1.6/(2M)] = 7.48 m/s