A small block slides with velocity $0.5\sqrt{\text{gr}}$ on the horizontal frictionless surface as shown in fig. Theblock leaves the surface at point C. The angle$\theta$ in the fig is:

At print ‘C’

$\text{N}=\text{mg}\cos \theta -\frac{{\text{mv}}_{\text{0}}^{\text{2}}}{\text{r}}$

$\text{mg}\cos \theta =\frac{{\text{mv}}_{\text{0}}^{\text{2}}}{\text{r}}\left(\because \text{N}=0\right)$

According to LCE

$\frac{1}{2}{\text{mv}}_{\text{0}}^{\text{2}}=\frac{1}{2}{\text{mu}}^{\text{2}}+\text{mgh}$

$\frac{1}{2}{\text{mv}}_{\text{0}}^{\text{2}}=\frac{1}{2}{\text{mu}}^{\text{2}}+\text{mg}r\left(1-\cos \theta \right)$

${\text{mv}}_{\text{0}}^{\text{2}}={\text{mu}}^{\text{2}}+2\text{mgr}\left(1-\cos \theta \right)$

${\text{mv}}_{\text{0}}^{\text{2}}={\text{mu}}^{\text{2}}+2\text{mgr}-2\text{mgr}\cos \theta$

$\text{mgr}\cos \theta =\text{m}{\left(0.5\sqrt{\text{gr}}\right)}^2+\text{2mgr}-2\text{mgr}\cos \theta$

$\text{gr}\cos \theta =\frac{1}{4}\text{gr}+\text{2gr}-2\text{gr}\cos \theta$

$3\text{g}\cos \theta =\frac{\text{9g}}{\text{4}}$

$\cos \theta =\text{3/4}$

$\theta ={\cos }^{-1}\left(\frac{3}{4}\right)$