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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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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 kg
Initial height ,h1= 12 metre
Final 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

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