A stone is projected with a velocity $20\sqrt{2}{\mathrm{ms}}^{-1}$ at an angle of 45⁰ to the horizontal. What will be the average velocity of stone during its motion from starting point to its maximum height? $\left(\mathrm{g}=10{\mathrm{ms}}^{-2}\right)$ 

Refer figure are when projectile is at A, then

$\mathrm{OC}=\frac{\mathrm{R}}{2}=\frac{1}{2}\frac{{\mathrm{u}}^2}{\mathrm{g}}\sin 2\theta =\frac{1}{2}\times \frac{(20\sqrt{2}{)}^2}{10}$$\sin \left(2\times {45}^{\circ }\right)$= 40

$\mathrm{AC}=\mathrm{H}=\frac{{\mathrm{u}}^2{\sin }^2\theta }{2\mathrm{g}}=\frac{(20\sqrt{2}{)}^2}{2\times 10}{\sin }^2{45}^{\circ }$=20m

$\therefore \text{ Displacement, }\mathrm{OA}=\sqrt{{\mathrm{OC}}^2+{\mathrm{CA}}^2}$

$=\sqrt{{40}^2+{20}^2}$

Time of projectile from O to A

$=\frac{1}{2}\left(\frac{2\mathrm{usin}\mathrm{\theta }}{\mathrm{g}}\right)=\frac{\mathrm{usin}\mathrm{\theta }}{2\mathrm{g}}=\frac{\left(20\sqrt{2}\right)\sin {45}^{\circ }}{10}=2s$

$\text{ Average velocity }=\frac{\text{ displacement }}{\text{ time }}$

$=\frac{\sqrt{{40}^2+{20}^2}}{2}=10\sqrt{5}{\mathrm{ms}}^{-1}$