A uniform solid spherical ball is rolling down a smooth declined plane from a height h. The velocity attained by the ball when it reaches the bottom of the incharged plane is v. If the ball is now thrown vertical upwards with the same velocity v, the maximum height to which the ball will rise is  [WB JEE 2014]

Total kinetic energy of a body which is rolling without slipping is given by

$K_{\text{total }}=K_{\text{rot }}+K_{\text{trans }}$

For solid spherical ball,  $I=\frac{2}{5}m{R}^2\text{ and }v=R\omega$   (Along the diameter)

where, R is radius of spherical ball.

So,    $K_{\text{total }}=\frac{1}{2}\left(\frac{2}{5}m{R}^2\right){\omega }^2+\frac{1}{2}m{R}^2{\omega }^2=\frac{7}{10}m{R}^2{\omega }^2=\frac{7}{10}m{v}^2$

Now, potential energy = kinetic energy

$\Rightarrow mgh=\frac{7}{10}m{v}^2\Rightarrow v^2=\frac{10}{7}gh$

For vertical projection,

$v^2=u^2+2g{h}^{\mathrm{\prime }}\Rightarrow \frac{10}{7}gh=0+2g{h}^{\mathrm{\prime }}\Rightarrow h^{\mathrm{\prime }}=\frac{5}{7}h$