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 5.4 m/s at an angle of 65° to the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor?

Question

Marlee Horne

Physics

22 November, 09:49

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 5.4 m/s at an angle of 65° to the vertical. If the ball rebounds with the same speed and angle, what was the impulse delivered to it by the floor?

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Answers (1)

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Jaslyn Sexton · Answer 1

2.74 N-s

Explanation:

m = mass of the basketball thrown towards the floor = 0.60 kg

v₀ = initial speed of the ball before hitting the floor = 5.4 m/s

θ = angle made by the direction of motion of the ball with the vertical before and after the collision = 65°

I = Impulse delivered to it by the floor

Impulse delivered to the ball by the floor is given as

I = 2 m v₀ Cosθ

Inserting the values

I = 2 (0.60) (5.4) Cos65

I = 2.74 N-s