Question

Two particles $A$ and $B$ are projected from point $O$ with equal speeds. They both hit the point $P$ of an inclined plane of inclination $15^{\circ }$. Particle $A$ is projected at an angle $30^{\circ }$ with inclined plane. If the ratio of time of flight of particles $A$ and $B$ is $1:\sqrt{n}$. The value of $n$ is _______.

Answer 1

Answer: 2

$\because y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

Or $0=u_{y}T-\frac{1}{2}\left(g\cos 30^{\circ }\right)T^2$
$\therefore T=\frac{2u_y}{g\cos 30^{\circ }}=\frac{4u_y}{10\sqrt{3}}$
$\therefore T_A=\frac{4u\sin 15^{\circ }}{10\sqrt{3}}$ ....(i)
and $T_B=\frac{4u\sin \alpha }{10\sqrt{3}}$
where, $\alpha$ is angle of projection of particle $B$,
$\because R=u_{x}T+\frac{1}{2}{a}_x{T}^2$....(ii)
Or $R=\left(u\cos 60^{\circ }\right)T_A-\frac{1}{2}g\sin 30^{\circ }{T}_A^2$
$=(u\cos \alpha )T_B-\frac{1}{2}g\sin 30^{\circ }{T}_B^2$
Or $u\cos 60^{\circ }\times \frac{4u\sin 60^{\circ }}{10\sqrt{3}}-\frac{1}{2}g\sin 30^{\circ }{T}_A^2$
$=(u\cos \alpha )T_B-\frac{1}{2}g\sin 30^{\circ }{T}_B^2$
Putting the value of $T_A$ and $T_B$,
$\alpha =90^{\circ }-15^{\circ }-30^{\circ }=45^{\circ }$
$\therefore \frac{T_A}{T_B}=\frac{4u\sin 30^{\circ }}{10\sqrt{3}}=\frac{4u\sin \alpha }{10\sqrt{3}}=\frac{\sin 30^{\circ }}{\sin \alpha }$
$=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}=1:\sqrt{n}$
$\therefore n=2$