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A ball is dropped from a height of 10m. If the energy of the ball  reduces by 40percent after striking ground, how high ball bounce

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H1 =10 m, v1 =?

Loss of energy =40% H2 =?

On striking the floor, K.E = P.E

½ mv12 = mg H1

V= ?(2gH1)= ?( 2x9.8x10)

V1= ?196= 14 m/s

As the body loses 40% of its energy, final energy E2 =( 60/100) of initial energy

i.e. mgH2 = (60/100)mgH1

H2 =3/5 H1= (3/5)X10= 6m

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