A particle is projected from ground with velocity $v$ at angle $\theta$ to horizontal. Average velocity of particle between point of projection and maximum height of projectile is-

The correct option is C $\frac{v}{2}\sqrt{1+3co{s}^2\theta }$
Range of the projectile motion$=R=\frac{2v^{2}cos\theta \sin \theta }{g}$

Maximum height attained by particle in the motion$=h=\frac{u_y^2}{2g}=\frac{u^{2}si{n}^2\theta }{2g}$

Total time taken by the particle to reach the topmost point $=t=\frac{2u_y}{g}=\frac{2usin\theta }{g}$

Average velocity of particle between two points is given as the displacement between the two points /time taken to reach that

point=$\frac{\sqrt{(\frac{R}{2}{)}^2+h^2}}{t}=\frac{v}{2}\sqrt{1+3co{s}^2\theta }$