A mass ‘m’ is released, with a horizontal speed v from the top of a smooth and fixed, hemispherical bowl, of radius r. The angle $\mathrm{\theta }$ w.r.t. the vertical where it leaves contact with the bowl is

Given: speed of block (horizontal) =v, mass of block =m, and radius of hemisphere =r

The block should be propelled horizontally with a velocity of v from the highest point, designated as P, and should travel a vertical distance of h before leaving the surface.
​${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{gh}={\mathrm{u}}^2+2\mathrm{gR}(1-\mathrm{cos\theta })\mathrm{\_\_\_\_\_}\left(1\right)$

$\mathrm{mgcos\theta }-\mathrm{N}=\frac{{\mathrm{mv}}^2}{\mathrm{R}}\mathrm{\_\_\_\_\_}\left(2\right)$

Here when the particle loses contact,

$\mathrm{N}=0\mathrm{\_\_\_}\left(3\right)$

from equations 1, 2 and 3 

${\mathrm{v}}^2=\mathrm{gRcos\theta }={\mathrm{v}}^2+2\mathrm{gR}(1-\mathrm{cos\theta })$

$3\mathrm{gRcos\theta }={\mathrm{v}}^2+2\mathrm{gR}$

$\mathrm{cos\theta }=\frac{{\mathrm{v}}^2}{3\mathrm{gR}}+\frac{2}{3}$

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

Hence the correct answer is $co{s}^{-1}\left(\frac{v^2}{3gR}+\frac{2}{3}\right).$