A basketball player makes a bounce pass by throwing the 0.60 kg basketball towards the floor at an angle,
Thetai= 65° to the vertical. The speed of the ball just before it hits the floor is 5.4 m/s. The ball rebounds with the same speed and angle
Thetaf= 65° relative to the vertical. What is magnitude of the impulse delivered by the floor to the ball?

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onyebuchinnaji onyebuchinnaji · Answer 1 · 2022-08-24T19:04:32+02:00

The magnitude of the impulse delivered by the floor to the ball is 6.48 N.s.

The impulse experienced by an object is the product of force and time of force action.

The impulse delivered by the floor to the ball is calculated from the change in momentum of the ball.

J = mvf - mvi

Jy = 0.6(5.4sin 65) - 0.6(-5.4sin 65)

Jy = 5.87 Ns

Jx = 0.6(5.4cos 65) - 0.6(-5.4sin 65)

Jx = 2.74 Ns

J = √(5.87² + 2.74²)

J = 6.48 N.s

Thus, the magnitude of the impulse delivered by the floor to the ball is 6.48 N.s.

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