Question

The velocity of a particle when it is at its greatest height is $\sqrt{\frac{2}{5}}$ of its velocity when it is at half its greatest height. The angle of projection of the particle is $\theta^{\circ}$. Find $\frac{\theta^{\circ}}{10}$.

Answer 1

Let particle be projected with velocity $u$ velocity at maximum height $=u \cos \theta$.
Initial velocity at half of maximum height
$v_{y}^{2}=u_{y}^{2}+2 a_{y} s_{y}$
$v_{y}^{2}=u^{2} \sin ^{2} \theta+2(-g) \frac{u^{2} \sin ^{2} \theta}{4 g}$
$v_{y}^{2}=\frac{u^{2} \sin ^{2} \theta}{2}$
$\Rightarrow v_{y}=\frac{u \sin \theta}{\sqrt{2}}$
Net velocity at half of maximum height
$=\sqrt{ u ^{2} \cos ^{2} \theta+\frac{ u ^{2} \sin ^{2} \theta}{2}}$
As per question,
$u \cos \theta=\sqrt{\frac{2}{5}} \sqrt{u^{2} \cos ^{2} \theta+\frac{u^{2} \sin ^{2} \theta}{2}}$
$5 u^{2} \cos ^{2} \theta=2 u^{2} \cos ^{2} \theta+u^{2} \sin ^{2} \theta$
$3 u^{2} \cos ^{2} \theta=2 u^{2} \sin ^{2} \theta$
$\tan ^{2} \theta=3 \Rightarrow \tan \theta=\sqrt{3}$
$\theta=60^{\circ}$