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^o. Particle A is projected at an angle 30^o with inclined plane. If the ratio of time of flight of particles A and B is $1:\sqrt{n}$. The value of n is ________.

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

$\begin{array}{l}\text{ Or }0=u_{y}T-\frac{1}{2}\left(g\cos {30}^{\circ }\right)T^2\\ \therefore T=\frac{2u_y}{g\cos {30}^{\circ }}=\frac{4u_y}{10\sqrt{3}}\\ \therefore T_A=\frac{4u\sin {15}^{\circ }}{10\sqrt{3}} ......\left(i\right)\\ \text{ and }{T}_B=\frac{4u\sin \alpha }{10\sqrt{3}}\end{array}$

where, $\alpha$ is angle of projection of particle B,

$\because R=u_{x}T+\frac{1}{2}{a}_x{T}^2 ......\left(ii\right)$

$\text{ Or }R=\left(u\cos {60}^{\circ }\right)T_A-\frac{1}{2}g\sin {30}^{\circ }{T}_A^2$

$\begin{array}{l}=(u\cos \alpha )T_B-\frac{1}{2}g\sin {30}^{\circ }{T}_B^2\\ \text{ Or }u\cos {60}^{\circ }\times \frac{4u\sin {60}^{\circ }}{10\sqrt{3}}-\frac{1}{2}g\sin {30}^{\circ }{T}_A^2\\ =(u\cos \alpha )T_B-\frac{1}{2}g\sin {30}^{\circ }{T}_B^2\end{array}$

Putting the value of T_A and T_B,

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\alpha ={90}^{\circ }-{15}^{\circ }-{30}^{\circ }={45}^{\circ }\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{T_A}{T_B}=\frac{\frac{4u\sin {30}^{\circ }}{10\sqrt{3}}}{\frac{4u\sin \alpha }{10\sqrt{3}}}=\frac{\sin {30}^{\circ }}{\sin \alpha }=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}}=1:\sqrt{n}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}n=2\end{array}$