A ball is dropped vertically from rest at a height of 12 m. After striking the ground, it bounces to a h eight of 9 m. What fraction of kinetic energy does it loose on striking the ground?

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Vibhor Maheshwari

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Let mass of the ball = m kgInitial height ,h1= 12 metreFinal height , h2= 9 metre Initial P.E = mgh1 = m × 9.8 × 12 Final P.E = m × g × h2 = m × 9.8 × 9 (Final P.E/ Initial P.E) ={ (m ×9.8 ×9) /(m ×9.8×12) } ...
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Let mass of the ball = m kgInitial height ,h1= 12 metreFinal height , h2= 9 metre Initial P.E = mgh1 = m × 9.8 × 12 Final P.E = m × g × h2 = m × 9.8 × 9 (Final P.E/ Initial P.E) ={ (m ×9.8 ×9) /(m ×9.8×12) } =3/4 Final P.E =[ 3 × Initial P.E] / 4 Loss in P.E is 1/4 of initial P.E % loss = (1/4)×100 = 25 read less
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