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

The correct option is D ${30}^{\circ }$

For vertically upward motion of a projectile,
$y=\left(u\sin \alpha \right)t-\frac{1}{2}{gt}^2$

where $\alpha$ is the angle of projection with the horizontal.
or $h=\left(u\sin \alpha \right)t-\frac{1}{2}{gt}^2$
or ${gt}^2-\left(2u\sin \alpha \right)t+2h=0$.....(i)
$\therefore t=\frac{2u\sin \alpha \pm \sqrt{\left(4u^2{sin}^2\alpha \right)-8gh}}{2g}$
If two roots of quadratic Eq. (i) are $t_1,t_2$, then
$t_1=\frac{2u\sin \alpha +\sqrt{4u^2{sin}^2\alpha -8gh}}{2g}$ .....(i)
and $t_2=\frac{2u\sin \alpha -\sqrt{\left(4u^2{sin}^2\alpha \right)-8gh}}{2g}$ ....(ii)
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 $t^2=t_1{t}_2$ .....(iii)

where total time of flight is given by $t=\frac{2u\sin \alpha }{g}$ ....(iv)

Putting (i), (ii) and (iv) in equation (iii), we get
${\left(\frac{2u\sin \alpha }{g}\right)}^2=\frac{{\left(2u\sin \alpha \right)}^2-(4u^2{sin}^2\alpha -8gh)}{4g^2}$
or $\frac{4u^2{sin}^2\alpha }{g^2}=\frac{8gh}{4g^2}$
or $2u^2{sin}^2\alpha =gh$
Given, $u=\sqrt{2gh}$
$\therefore 2\left(2gh\right){sin}^2\alpha =gh$
or ${sin}^2\alpha =\frac{1}{4}$
or $\sin \alpha =\frac{1}{2}$
$\therefore \alpha ={30}^{\circ }$