When the ratio of the minimum velocity to the maximum velocity is $\frac{1}{\sqrt{2}}$ in a projectile motion, the angle of projection of projectile is

The correct option is D ${45}^{\circ }$
Let $u$ be the initial velocity of a projectile
We know that the velocity of projectile along the trajectory is minimum when its vertical component vanishes and only horizontal component remains (i.e. at the top).
$\Rightarrow v_{min}=u\cos \theta$
$\Rightarrow$ Ratio of minimum to maximum velocity $=\frac{u\cos \theta }{u}=\frac{1}{\sqrt{2}}$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta =45°$