A coin is sliding down on a smooth hemi- spherical surface of radius R. The height from the bottom, where it looses contact with the surface is

The coin will leave contact with the  surface,
Where N = 0
 $\Rightarrow mg\cos \theta -\frac{m{v}^2}{R}=0$  
 $mg\cos \theta =\frac{m{v}^2}{R}\Rightarrow \cos \theta =\frac{v^2}{Rg}$
According to L.C of energy mgy + $\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2$ 
Body leaves  $(v_0=0)$ where y=R(1-cosθ)
$\cos \theta =\frac{2}{3}fromfig\cos \theta =\frac{h}{R}\Rightarrow h=\frac{2R}{3}$