From the figure, \(mg = R\sin \theta \) …\((i)\)
\(\;\frac{{m{v^2}}}{r} = R\cos \theta \) …\((ii)\)
From equation \((i)\) and \((ii)\) we get
\(\tan \theta = \frac{{rg}}{{{v^2}}}\) but \(\tan \theta = \frac{r}{h}\)
\(h = \frac{{{v^2}}}{g} = \frac{{{{(0.5)}^2}}}{{10}} = 0.025m = 2.5\,cm\)