The speed of a projectile when at its greatest height is $\sqrt{\frac{2}{5}}$  of its speed when at half of  its greatest height. Angle of projection is

$V_x=u\cos \theta$

$V_y=\sqrt{u^2{\sin }^2\theta -2gh}$

Here $V_y=\sqrt{u^2{\sin }^2\theta -2g\left(\frac{h}{2}\right)}$

$=\sqrt{u^2{\sin }^2\theta -\frac{2g}{2}\left(\frac{u^2{\sin }^2\theta }{2g}\right)}$

$=\sqrt{\frac{u^2{\sin }^2\theta }{2}}$

The velocity at that point be

$V_m=\sqrt{V_x^2+V_y^2}=\sqrt{u^2{\cos }^2\theta +\frac{u^2{\sin }^2\theta }{2}}$

Velocity at heighest point is $V_H=V_x=u\cos \theta$

Given $V_H=\sqrt{\frac{2}{5}}{V}_m$

$V_H^2=\frac{2}{5}{V}_m^2$

$u^2{\cos }^2\theta =\frac{2}{5}\left(\frac{2u^2{\cos }^2\theta +u^2{\sin }^2\theta }{2}\right)$

${\tan }^2\theta =3$

$\therefore \theta ={60}^0$