Question

A particle is projected with velocity $\sqrt{2gh}$ , such that it just crosses two walls of height $h$ and separated by $h$ . Find the angle of projection.

• A. $15^{\circ }$
• B. $75^{\circ }$
• C. $60^{\circ }$
• D. $30^{\circ }$

Answer 1

Answer: (D)

For vertically upward motion of a projectile,
$y=(u\sin \alpha )t-\frac{1}{2}g{t}^2$
or $h=(u\sin \alpha )t-\frac{1}{2}g{t}^2$
or $g{t}^2-(2u\sin \alpha )t+2h=0$ ...(i)
$\therefore t_1=\frac{2u\sin \alpha \pm \sqrt{\left(4u^2{\sin }^2\alpha \right)-8gh}}{2g}$
If two roots of quadratic Eq. (i) are $t_1,t_2$, then
$t=\frac{2u\sin \alpha +\sqrt{(4u^2{\sin }^2\alpha -8gh}}{2g}$
and $t_2=\frac{2u\sin \alpha -\sqrt{\left(4u^2{\sin }^2\alpha \right)-8gh}}{2g}$
If particle crosses the walls at times $t_1$ and $t_2$ respectively, then time of flight $t$ is $t=\sqrt{t_1{t}_2}$
or $t^2=t_1{t}_2$
$\therefore {\left(\frac{2u\sin \alpha }{g}\right)}^2=\frac{(2u\sin \alpha {)}^2-\left(4u^2{\sin }^2\alpha -8gh\right)}{4g^2}$
or $\frac{4u^2{\sin }^2\alpha }{g^2}=\frac{8gh}{4g^2}$
or $2u^2{\sin }^2\alpha =gh$
Given, $u=\sqrt{2gh}$
$\therefore 2(2gh){\sin }^2\alpha =gh$
or ${\sin }^2\alpha =\frac{1}{4}$
or $\sin \alpha =\frac{1}{2}$
$\therefore \alpha =30^{\circ }$