The potential energy of a projectile at maximum height is 3/4 times kinetic energy of projection. Its angle of projection is

The P.E of a projectile at maximum height

$={\text{mgh}}_{\max }=\text{mg}\frac{{\text{u}}^{\text{2}}{\sin }^2\theta }{2g}$

$=\frac{{\text{mu}}^{\text{2}}{\sin }^2\theta }{2}$

K.E of projection $=\frac{1}{2}{\text{mu}}^{\text{2}}$

given that$\text{PE}=\frac{3}{4}\text{K}\text{.E}$

$\frac{{\text{mu}}^2{\sin }^2\theta }{2}=\frac{3}{4}\frac{1}{2}{\text{mu}}^{\text{2}}$

$\sin \theta =\frac{\sqrt{3}}{2}$

hence$\theta =60°$