When an object is shot from the bottom of a long smooth inclined plane kept at an angle $60^{\circ}$

Question

When an object is shot from the bottom of a long smooth inclined plane kept at an angle $60^{\circ}$ with horizontal, it can travel a distance $x_{1}$ along the plane. But when the inclination is decreased to $30^{\circ}$ and the same object is shot with the same velocity, it can travel $x_{2}$ distance. Then $x_{1}: x_{2}$ will be\

Professor PC Thomas Classes · Answer

(d) : $v^{2}-u^{2}=2 a x$ For case I: $v^{2}=u^{2}-2\left(g \sin 	heta_{1}ight) x_{1} \quad[\because a=-g \sin 	heta]$\ $x_{1}=\frac{u^{2}}{2 g \sin 	heta_{1}}$\ $[\because v=0]$\ For case II :\ $v^{2}=u^{2}-\left(2 g \sin 	heta_{2}ight) x_{2}$\ $x_{2}=\frac{u^{2}}{2 g \sin 	heta_{2}}$\ $	herefore \frac{x_{1}}{x_{2}}=\frac{\sin 	heta_{2}}{\sin 	heta_{1}}=\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1}{\sqrt{3}}$\