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

Answer: 6.00

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