Q. When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with horizontal, it can travel a distance $x_1$ along the plane. But when the inclination is decreased to 30° and the same object the shot with the same velocity, it can travel $x_2$ distance. Then $x_1 : x_2$ will be
A) $1 : \sqrt{2}$
B) $\sqrt{2} : 1$
C) $1 : \sqrt{3}$
D) $1 : 2\sqrt{3}$
Explanation
[Solution] $x = \frac{u^2}{2g\sin\theta}$, so $x_1 : x_2 = \sin 30^\circ : \sin 60^\circ = 1 : \sqrt{3}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.