A small spherical ball is released from a height h on the rough...

By the principle of conservation of mechanical energy, the loss in potential energy is equal to the gain in total kinetic energy.

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

For a solid spherical ball, the moment of inertia about its center of mass is $I = \dfrac{2}{5}mr^2$.

Since the ball rolls without slipping, the linear speed $v$ and angular speed $\omega$ are related by $v = \omega r$.

Substituting $I$ and $\omega$ into the energy equation:

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2} \left(\dfrac{2}{5}mr^2\right) \left(\dfrac{v}{r}\right)^2$

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{5}mv^2$

$mgh = \dfrac{7}{10}mv^2$

$v^2 = \dfrac{10gh}{7}$

$v = \sqrt{\dfrac{10gh}{7}}$