Average velocity of a particle in projectile motion between its starting point and the highest point of its trajectory is (projection speed =$u$, angle of projection from horizontal =$\theta$)

Total displacement between the initial point and the final point is:

$$\begin{gathered} d=\sqrt{h^2+{\left(\frac{R}{2}\right)}^2}=\sqrt{\frac{u^4{\sin }^4\theta }{4g^2}+\frac{u^4{\sin }^{2}2\theta }{4g^2}}=\frac{u^2\sin \theta }{2g}\sqrt{{\sin }^2\theta +4{\cos }^2\theta } \\ \therefore d=\frac{u^2\sin \theta }{2g}\sqrt{1+3{\cos }^2\theta } \end{gathered}$$

$$\begin{gathered} \text{ Time taken to reach the height, }t=\frac{T}{2}=\frac{2u\sin \theta }{2g}=\frac{u\sin \theta }{g} \\ \text{Average velocity }=\frac{d}{t}=\frac{\frac{u^2\sin \theta }{2g}\sqrt{1+3{\cos }^2\theta }}{\frac{u\sin \theta }{g}}=\frac{u}{2}\sqrt{1+3{\cos }^2\theta } \end{gathered}$$