A particle is projected from the ground with an inital speed of $v$ at an angle $\theta$ with horizontal. The average velocity of the particle between its point of projection and highest point of trajectory is

The correct option is D $\frac{v}{2}\sqrt{1+3{\cos }^2\theta }$
Draw a rough diagram of the given problem



Find average velocity of the particle
We know,
$\text{Average velocity}=\frac{\text{total displacement}}{\text{total time}}$
From the figure total displacement of the particle is
$=\sqrt{H^2+\frac{R^2}{4}}$
and in this situation time taken by the particle is $\frac{T}{2}$
Then,
$v_{av}$$=\frac{\sqrt{H^2+\frac{R^2}{4}}}{\frac{T}{2}}$$\dots \left(i\right)$

As we know,
$H=\text{Maximum height}=\frac{v^2{\sin }^2\theta }{2g}$
$R=\text{Range}=\frac{v^2\sin 2\theta }{g}$
$T=\text{Time of flight}=\frac{2u\sin \theta }{g}$
substituting the value in equation $\left(i\right)$

$v_{av}=\frac{\sqrt{\left(\frac{v^4{\sin }^4\theta }{4g^2}\right)+\frac{1}{4}\left(\frac{4v^4{\sin }^2\theta {\cos }^2\theta }{g^2}\right)}}{\frac{2v\sin \theta }{2g}}$
$v_{av}=\left(\frac{v^2\sin \theta }{2g}\right)\left(\frac{\sqrt{1+3{\cos }^2\theta }}{\frac{v\sin \theta }{g}}\right)$
$\therefore v_{av}=\frac{v}{2}\sqrt{1+3{\cos }^2\theta }$