Solve the following :

From figure,

$\left(m v^{2}\right) / R=m g \cos \theta$

$v=(\sqrt{g} R \cos \theta) \ldots \ldots 1$

and

Change in K.E. = Work done

$\frac{1}{2} m v^{2}-0=m g(R-R \cos \theta)$

$v=(\sqrt{2} g R[1-\cos \theta]) \ldots . .2$

from 1 and 2

$g R \cos \theta=2 g R(1-\cos \theta)$

$\theta=\cos ^{-1}\left(\frac{2}{3}\right)$