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

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

Or $0=u_{y} T-\frac{1}{2}\left(g \cos 30^{\circ}\right) T^{2}$
$\therefore T=\frac{2 u_{y}}{g \cos 30^{\circ}}= \frac{4u_y}{10\sqrt{3}}$
$\therefore T_{A}=\frac{4 u \sin 15^{\circ}}{10 \sqrt{3}}$ ....(i)
and $T_{B}=\frac{4 u \sin \alpha}{10 \sqrt{3}}$
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{4 u \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}$
$\therefore \frac{T_{A}}{T_{B}}=\frac{4 u \sin 30^{\circ}}{10 \sqrt{3}}=\frac{4 u \sin \alpha}{10 \sqrt{3}}=\frac{\sin 30^{\circ}}{\sin \alpha}$
$=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}=1: \sqrt{n}$
$\therefore n=2$