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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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(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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$A$ 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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Two bodies of equal masses move with the uniform velocities $v$ and $3v$ respectively. Find the ratio of their kinetic energies.

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$(a)$ Define potential energy. Write an expression for potential energy of an object of mass $m$ raised through a height $h$.
$(b)$ Find the energy possessed by an object of mass $10 \, kg$ when it is raised to a height of $6 \, m$ above the ground. (Given $g = 9.8 \, m s^{-2}$)