A uniform solid spherical ball is rolling down an inclined plane from a height h. The velocity attained by the ball when it reaches the bottom of the inclined plane is v. If the ball is now thrown vertically upwards with the same velocity v, the maximum height to which the ball will rise is

c) Total kinetic energy of a body which is rolling without slipping is given by
 $K_{total}=K_{rot}+K_{trans}$
For solid spherical ball $I=\frac{2}{5}m{R}^2$  and  $v=R\omega$ (Along the diameter)
Where, R is radius of spherical ball.
 $So,K_{total}=\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\omega }^2+\frac{1}{2}m{R}^2{\omega }^2=\frac{7}{10}m{v}^2$
Now potential energy = kinetic energy 
$\Rightarrow mgh=\frac{7}{10}m{v}^2\Rightarrow v^2=\frac{10}{7}gh$ ……………..(i)
For vertical projection,  $v^2=u^2+2gh\text{'}\Rightarrow \frac{10}{7}gh=o+2gh\text{'}\Rightarrow h\text{'}=\frac{5}{7}h$