The loss in gravitational potential energy is equal to the gain in translational kinetic energy and rotational kinetic energy for the case of maximum velocity of sphere when it reaches ground. 

$mgh=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$, here the relation, $\omega =\frac{v}{r}$ and $I=\frac{2}{5}m{r}^2$, it means, 

$$\begin{gathered} mgh=\frac{1}{2}\times \frac{2}{5}m{r}^2\times {\frac{v}{r^2}}^2+\frac{1}{2}m{v}^2=\frac{1}{2}\left(\frac{2}{5}m{v}^2+m{v}^2\right) \\ \Rightarrow gh=\frac{1}{2}\left(\frac{7}{5}{v}^2\right) \\ \Rightarrow v=\sqrt{\frac{10}{7}gh} \end{gathered}$$

Only depends on height and acceleration due to gravity.