A 0.400-kg ball is dropped from rest at a point 1.80 m above the floor. The ball rebounds straight upward to a height of 0.730 m. What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

Answer 1

Get the velocity before floor impact from the height H from which it is dropped. Call it V1 (positive down)

V1 = sqrt(2H/g)

Get the velocity just after floor impact from the height H' to which it rises. Call that V2 (positive up)

V2 = sqrt(2H'/g)

The momentum change, which is the impulse, is
M*(V1 + V2)

The impulse on the ball is up, since the momentum increases in the upward direction.