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

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SUMMARY

The discussion centers on calculating the impulse delivered to a 0.60 kg basketball during a bounce pass. The ball strikes the floor at a speed of 5.4 m/s at a 65-degree angle to the vertical and rebounds elastically with the same speed and angle. The impulse is determined using the equation I = ΔP, where the change in vertical velocity (ΔVy) is calculated. It is confirmed that the impulse is directed vertically upward, as the horizontal component of velocity (ΔVx) is zero, indicating that the angle of collision does not affect the direction of the impulse.

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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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