Question

The velocity of a projectile when it is at the greatest height is $\sqrt{2/5}$ times its velocity when it is at half of its greatest height. Determine its angle of projection.

Answer 1

Suppose the particle is projected with velocity u at an
angle $\theta$ with the horizontal. Horizontal
component of its velocity at all height will be u $\cos \theta .$
At the greatest height, the vertical component of velocity is zero, so the resultant velocity is
$v_1=u\cos \theta$
At half the greatest height during upward motion,
$y=h/2,$
$a_y=-g,$
$u_y=u\sin \theta$
Using $v_y^2-u_y^2=2a_{y}y$
we get, $v_y^2-u^2{\sin }^2\theta =2(-g)\frac{h}{2}$
or $v_y^2=u^2{\sin }^2\theta -g\times \frac{u^2{\sin }^2\theta }{2g}=\frac{u^2{\sin }^2\theta }{2}$ $[\because h=\frac{u^2{\sin }^2\theta }{2g}]$
or $v_y=\frac{u\sin \theta }{\sqrt{2}}$
Hence, resultant velocity at half of the greatest height is
$v_2=\sqrt{v_x^2+v_y^2}$
$=\sqrt{u^2{\cos }^2\theta +\frac{u^2{\sin }^2\theta }{2}}$
Given, $\frac{v_1}{v_2}=\sqrt{\frac{2}{5}}$
$\therefore \frac{v_1^2}{v_2^2}=\frac{u^2{\cos }^2\theta }{u^2{\cos }^2\theta +\frac{u^2{\sin }^2\theta }{2}}=\frac{2}{5}$
or $\frac{1}{1+\frac{1}{2}{\tan }^2\theta }=\frac{2}{5}$
or $2+{\tan }^2\theta =5or{\tan }^2\theta =3$
or $\tan \theta =\sqrt{3}$
$\therefore \theta ={60}^{\circ }$