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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?

A basketball player makes a bounce pass by throwing the 060 kg basketball towards the floor at an angle Thetai 65 to the vertical The speed of the ball just bef class=

Respuesta :

onyebuchinnaji

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

What is impulse?

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

Impulse delivered by the floor to the ball

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.

Learn more about impulse here: https://brainly.com/question/25700778

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