What is the direction of impulse delivered to a bouncing ball?

AI Thread Summary
The discussion focuses on calculating the impulse delivered to a basketball during a bounce pass. The player throws a 0.60 kg ball at a speed of 5.4 m/s and an angle of 65 degrees to the vertical, and it rebounds elastically with the same speed and angle. It is confirmed that the impulse is directed vertically upward since the horizontal component of velocity change (ΔVx) is zero, meaning the angle of collision does not affect the direction of impulse. The magnitude of the impulse, however, is influenced by the angle of impact. Overall, the impulse delivered by the floor is vertical, with its magnitude depending on the collision angle.
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Homework Statement


To make a bounce pass, a player throws a 0.60 kg basketball towards the floor. the ball hits the floor with a speed of 5.4m/s at an angle of 65o to the vertical. The ball rebounds elastically with the same speed and angle. Calculate impulse (magnitude and direction) delivered to the ball by the floor.

Homework Equations


if ^ is delta
I=^P

The Attempt at a Solution



^Vx=0

^Vy=Vyf-Vyi=

5.4m/s(cos(65o)+5.4m/s(cos(65o)

Impulse= ^P

I just want to confirm that the impulse delivered to the ball will all be directed vertically upward because ^Vx=0. So the angle of collision will have no effect on the direction of impulse in this case.
 
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The direction remains the same, but the magnitude depends on the angle.
 
Thank you
 
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