\(\frac{1}{2}m{v^2} + \frac{1}{2} \cdot \frac{1}{2}m{R^2} \cdot \frac{{{V^2}}}{{{R^2}}} = mg{h_{sph}}\)
For solid cylinder
\(\frac{1}{2}m{V^2} + \frac{1}{2} \cdot \frac{1}{2}m{R^2} \cdot \frac{{{V^2}}}{{{R^2}}} = mg{h_{cyl}}\)
\( \Rightarrow \frac{{{h_{sph}}}}{{{h_{cyl}}}} = \frac{{7/5}}{{3/2}} = \frac{{14}}{{15}}\)