To make a bounce pass, a player throws a 0.60-kg basketball toward the floor. The ball hits the floor with a speed of 5.5 m/s at an angle of 59° from the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor?

Answer 1

The impulse delivered by the floor equals the momentum change in the vertical direction.

That is equal to 2*M*Vcos55

The horizontal velocity component does not change during the bounce.

Answer 2

In this case, I=2MVsin31