A ball at rest is dropped from a height of 12 | vedclass

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(C) Let the initial height be $H = 12 ~m$. The potential energy of the ball at this height is $PE_1 = mgH$.When the ball strikes the ground,its kineti

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$9$

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(C) Let the initial height be $H = 12 ~m$. The potential energy of the ball at this height is $PE_1 = mgH$.When the ball strikes the ground,its kinetic energy just before impact is $KE_1 = mgH$.The ball loses $25 \%$ of its kinetic energy,so the remaining kinetic energy is $KE_2 = KE_1 - 0.25 KE_1 = 0.75 KE_1$.The ball bounces back to a height '$h$',so its potential energy at the maximum height reached is $PE_2 = mgh$.By the law of conservation of energy,the kinetic energy after the bounce is equal to the potential energy at the new height: $KE_2 = PE_2$.Therefore,$mgh = 0.75 mgH$.Canceling $mg$ from both sides,we get $h = 0.75 H$.Substituting $H = 12 ~m$,we get $h = 0.75 	imes 12 = 9 ~m$.

$A$ ball at rest is dropped from a height of $12 ~m$. It loses $25 \%$ of its kinetic energy on striking the ground and bounces back to a height '$h$'. Then the value of '$h$' is: (in $~m$)

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