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A rubber ball of mass $m$ is released from rest at height $h$ above the floor. After its first bounce, it rises to 90$\%$ of its original height. What impulse (magnitude and direction) does the floor exert on this ball during its first bounce? Express your answer in terms of the variables $m$ and $h$ .

2.76$m \sqrt{h}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

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here for the solution. The explanation is that when the ball dropped from a height edge, the velocity V at it will hit. The ground is given by that half. MV is critical to MGH the wretches, The hygiene is the dude accession due to gravity and Mr Mosque with the velocity so it will be equal to under route to G H now. So initial velocity dated vehicle to undertook to GH After first bounds, it reached height 90% off edge. So the velocity just after hitting them ground that the F net is final, velocity will be called to undergo to be multiplied by 0.9 edge and this will be called to under owed 1.8 gs. Now we a factor is equal to minus were director. So the impulse get Jake will too integrated, Affected DT So it will be m in to get deep by DT DT So Jay factor will be called, um limit v I to v f DV not. It will be m v f factor minus the director M v Air Palace v I, and kept. And here by substituting the value, it will be M. Underwood one point at G H Plus under route to GH and kept and help again. We simplified this. We get 2.76 am under load GH and cap and develop Final answer. Where and cap is the unit normal Toe the surface along the direction we affect er So this is the explanation of the ancestor by step. Please, God.

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