A solid spherical ball rolls on a horizontal surface at $10 \mathrm{m} {\mathrm{s}}^{-1}$ and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is $11 \mathrm{kg}$ and frictional losses are negligible, the value of $h$, where the ball stops and starts rolling down the inclination is (Assume $\mathrm{g}=10 \mathrm{m} {\mathrm{s}}^{-2}$)

Given,

Initial velocity of solid spherical ball is $v=10 \mathrm{m} {\mathrm{s}}^{-1}$

Mass of ball rolling on the surface is $m=11 \mathrm{kg}$

Applying work energy theorem,

$W=∆KE$

$W_g=K{E}_f-K{E}_i ...\left(1\right)$

Now, initial kinetic energy of spherical ball is $K_i=\frac{1}{2}m{v}^2\left(1+\frac{k^2}{R^2}\right)$

And final kinetic energy is given by

$K_f=0$

Work done due to gravity is 

$W_g=-mgh$

Substitute the values in equation (1) we get

$\Rightarrow -mgh=0-\frac{1}{2}\times 11\times {10}^2\left(1+\frac{2}{5}\right)$

Where, $\frac{k^2}{R^2}=\frac{2}{5}$

$\Rightarrow h=\frac{350}{50}=7 \mathrm{m}$