Work done by friction during pure rolling is zero therefore, we can apply conservation of mechanical energy.

In pure rolling $\omega =\frac{v}{R}$

For a body having rotational and translational kinetic energy,

$$\begin{gathered} \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mgh \\ \Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{R}^2}{2}{\left(\frac{v}{R}\right)}^2=mgh \\ \Rightarrow v=\sqrt{\frac{4gh}{3}}=\sqrt{\frac{4\times 10\times 30}{3}}=20 \mathrm{m} {\mathrm{s}}^{-1} \end{gathered}$$