A hollow spherical ball of uniform density rolls up a curved...

The total kinetic energy of the ball at the bottom of the curve can be calculated as follows:

$\begin{cases}K & = & \dfrac{1}{2}M{v}^{2} + \dfrac{1}{2} \cdot \dfrac{2}{3}M{R}^{2}{\omega}^{2} \\ & = & \dfrac{1}{2}M{v}^{2} + \dfrac{1}{3}M{\left(R\omega\right)}^{2} \\ & = & \dfrac{1}{2}M{v}^{2} + \dfrac{1}{3}M{v}^{2}\left[∵ v = \omegaR\right] \\ & = & \dfrac{5}{6}M{v}^{2} . . . \left(1\right)\end{cases}$

The formula to calculate the potential energy acquired by the ball when it attains the maximum height is given by

$U = Mgh . . . \left(2\right)$

Equate equation (1) and (2) and simplify to obtain the required height.

$\begin{cases}Mgh & = & \dfrac{5}{6}M{v}^{2} \\ & ⇒ & h = \dfrac{5{v}^{2}}{6g} . . . \left(3\right)\end{cases}$

Substitute the values of the known parameters into equation (3) to calculate the required height attained by the ball.
$\begin{cases}h & = & \dfrac{5 \times {\left(3m{s}^{- 1}\right)}^{2}}{6 \times 10m{s}^{- 2}} \\ & = & 0 . 75m \times \dfrac{100cm}{1m} \\ & = & 75cm\end{cases}$