A sphere begins rolling down an incline with an inclination θ ....

By conservation of mechanical energy, the loss in potential energy equals the gain in kinetic energy.

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

For a solid sphere, the moment of inertia is $I = \dfrac{2}{5}mR^2$.

For pure rolling, $v = \omega R$.

The vertical distance fallen is $h = l\sin\theta$.

Substituting these values into the energy equation:

$mgl\sin\theta = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mR^2\right)\left(\dfrac{v}{R}\right)^2$

$mgl\sin\theta = \dfrac{1}{2}mv^2 + \dfrac{1}{5}mv^2$

$mgl\sin\theta = \dfrac{7}{10}mv^2$

$v^2 = \dfrac{10}{7}gl\sin\theta$

$v = \sqrt{\dfrac{10}{7}gl\sin\theta}$