The speed of a projectile at its highest point is v₁ and at the point half the maximum height is v₂. If $\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\sqrt{\frac{2}{5}}$, then find the angle of projection.

$\begin{array}{l}{\mathrm{v}}_1={\mathrm{v}}_2\cos \mathrm{\beta }=\mathrm{ucos}\mathrm{\theta }\\ 0={\left({\mathrm{v}}_2\sin \mathrm{\beta }\right)}^2-2\mathrm{g}(\mathrm{H}/2)\\ \Rightarrow {\mathrm{v}}_1^2={\mathrm{v}}_2^2-\mathrm{gH}\end{array}$

$\text{ Also }\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\sqrt{\frac{2}{5}}$

$\text{ From above }{\mathrm{v}}_1=\sqrt{\frac{2\mathrm{gH}}{3}},{\mathrm{v}}_2=\sqrt{\frac{5\mathrm{gH}}{3}}$

$\begin{array}{l} \mathrm{H}=\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}}\Rightarrow {\sin }^2\mathrm{\theta }=\frac{2{\mathrm{gHcos}}^2\mathrm{\theta }}{{\mathrm{v}}_1^2}\\ \mathrm{or} {\tan }^2\mathrm{\theta }=\frac{2\mathrm{gH}}{{\mathrm{v}}_1^2}\text{ or }{\tan }^2\mathrm{\theta }=\frac{2\mathrm{gH}\times 3}{2\mathrm{gH}}\\ \mathrm{or} \tan \mathrm{\theta }=\sqrt{3}\text{ or }\mathrm{\theta }={60}^{\circ }\end{array}$