A ball is dropped from a height of 10 , m. If | vedclass

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(B) Initial potential energy of the ball at height $h = 10 \, m$ is $E = mgh$.After striking the ground,the energy reduces by $40 \%$,which means the

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$6$

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(B) Initial potential energy of the ball at height $h = 10 \, m$ is $E = mgh$.After striking the ground,the energy reduces by $40 \%$,which means the ball retains $60 \%$ of its initial energy.Energy retained $= 0.60 	imes mgh$.When the ball bounces back to a new height $h'$,its potential energy at that height will be $mgh'$.Equating the retained energy to the new potential energy: $mgh' = 0.60 	imes mgh$.Canceling $mg$ from both sides,we get $h' = 0.60 	imes h$.Substituting $h = 10 \, m$,we get $h' = 0.60 	imes 10 \, m = 6 \, m$.Therefore,the ball bounces back to a height of $6 \, m$.

$A$ ball is dropped from a height of $10 \, m$. If the energy of the ball reduces by $40 \%$ after striking the ground,how much high can the ball bounce back (in $, m$)? $(g = 10 \, m s^{-2})$

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