A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?

Question

answer in kg·m/s

Answer 1

figure the velocity at impact, and leaving impact.

at impact, velocity= sqrt 2gh
returning, velocity= sqrt 2gh*.6 opposite direction

change velocity= Vi-Vf=sqrt2gh+sqrt(2gh*.6)

impulse= mass*changevelocity, upward direction.

Answer 2

8.78