Thank you for reporting, we will resolve it shortly
Q.
The speed of a projectile when it is at its greatest height is $\sqrt{2 / 5}$ times its speed at half the maximum height. What is its angle of projection?
Motion in a Plane
Solution:
Maximum height, $H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
or $gH=\frac{u^{2} \sin ^{2} \theta}{2}\,\,\,\,\,\,\dots(i)$
Velocity at the highest point, $v_{H}=u \cos \theta$
Let $v_{x}, v_{y}$ be the horizontal and vertical velocity of projectile at height $\frac{H}{2} .$ Then, $v_{x}=u \cos \theta$ and
$v_{y}^{2} =u^{2} \sin ^{2} \theta-2 g \times \frac{H}{2}=u^{2} \sin ^{2} \theta-g H $
$=u^{2} \sin ^{2} \theta-\frac{u^{2} \sin ^{2} \theta}{2}=\frac{u^{2} \sin ^{2} \theta}{2}$ (Using(i))
$\therefore $ Net velocity at height $\frac{H}{2}=\left(v_{x}^{2}+v_{y}^{2}\right)^{1 / 2}$
As per question
$\sqrt{\frac{2}{5}}\left(v_{x}^{2}+v_{y}^{2}\right)^{1 / 2}=v_{H}$ or
$ \frac{2}{5}\left(v_{x}^{2}+v_{y}^{2}\right)=v_{H}^{2} $
or $ \frac{2}{5}\left[u^{2} \cos ^{2} \theta+\frac{u^{2}}{2} \sin ^{2} \theta\right]=u^{2} \cos ^{2} \theta $
or $ \sin ^{2} \theta=3 \cos ^{2} \theta $ or $ \sin \theta=\sqrt{3} \cos \theta $
or $ \tan \theta=\sqrt{3}=\tan 60^{\circ} $ or $ \theta=60^{\circ}$