91Ó°ÊÓ

\({\rm{Height}}\;{\rm{at}}\;{\rm{rest}} = {H_0}\)

The total work done by the friction corresponds to the path of the object.

Let M is the mass and R is the radius of the ball.

Let v is the linear speed and \(\omega \) is the angular speed.

Apply conservation of energy,

\(\begin{aligned}{}\frac{1}{2}M{v^2} + \frac{1}{2}I{\omega ^2} &= Mg{H_0}\\\frac{1}{2}M{v^2} + \frac{1}{2} \times \frac{2}{3}M{R^2} \times {\omega ^2} &= Mg{H_0}\;\;\;\;\;\;\;\;\;\left( {I = \frac{2}{3}M{R^2}} \right)\\\frac{1}{2}M{v^2} + \frac{1}{3} \times M{\left( {R \times \omega } \right)^2} &= Mg{H_0}\;\\\frac{1}{2}M{v^2} + \frac{1}{3} \times M{\left( v \right)^2} &= Mg{H_0}\;\\\frac{5}{6}M{v^2} &= Mg{H_0}\;\\{v^2} &= \frac{{6g{H_0}}}{5}\end{aligned}\)

Maximum height reached by the ball on smooth surface,

\(\begin{aligned}{}{H_{\max }} &= \frac{{{v^2}}}{{2g}}\\{H_{\max }} &= \frac{{\frac{{6g{H_0}}}{5}}}{{2g}}\\ &= 3{H_0}/5\end{aligned}\)

Hence, the ball will go \(3{H_0}/5\) high.