A body is projected from the ground with a velocity $u$ at an angle $\theta$ with the horizontal. The average velocity of the body between its point of projection and the highest point of its trajectory is

A is the highest point on the trajectory. 

$$\begin{gathered} Average velocity =\frac{displacement OA}{time taken} \\ = \frac{\sqrt{h_{max}^2+{\displaystyle \frac{R^2}{4}}}}{{\displaystyle \frac{1}{2}}{t}_f} ......................\left(1\right) \\ where h_{max} = \frac{u^2{\sin }^2\theta }{2g}, \\ R = \frac{u^2{\sin }^{2}2\theta }{g}, \\ and t_{f }=\frac{2u \sin \theta }{g} \end{gathered}$$

Using these in Eq. (1) and simplifying, we get

 Average velocity  $=\frac{u}{2}{\left(1+3{\cos }^2\theta \right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$