Question
Asked Dec 18, 2019
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A ball of mass 0.150 kg is dropped from rest from a height
of 1.25 m. It rebounds from the floor to reach a height of
0.960 m. What impulse was given to the ball by the floor?

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The height from which the ball is dropped is 1.25m, and the rebounded height is 0.960m. The expression for velocity of ball just before the impact is V = -Vv + 2aAh Since the ball is under free fall , magnitude of acceleration is due to acceleration due to gravity. And the initial velocity of the ball is zero. 0- 2gAh +2(-9.8m/s*)(-1.25m) v, = - 10 +20 [1] V, = -4.95m/s The expression for rebound velocity of the ball is given by - 2aAh Since the ball is rebounding, magnitude of acceleration of the ball is due to acceleration due to gravity. And the final velocity of the ball is zero since the ball is at rest finally. v, = - 2gAh o- 2(9.8m/s*)(0.960m) V, %3D [2] v, = 4.34m/s %3D

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Science

Physics

Newtons Laws of Motion

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