The linear speed of a uniform spherical shell after rolling down an inclined plane of vertical height h from rest, is
 

Using the law of conservation of energy $mgh=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2\text{ but }{I}_{\text{shell }}=\frac{2}{3}m{R}^2$
 

$mgh=\frac{1}{2}\times \frac{2}{3}m{R}^2\cdot \frac{v^2}{R^2}+\frac{1}{2}m{V}^2\Rightarrow mgh=\frac{1}{3}m{v}^2+\frac{1}{2}m{v}^2;mgh=\frac{5}{6}m{v}^2\Rightarrow v^2=\frac{6gh}{5}$