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:

• A. $1:\sqrt{3}$
• B. $1: 2\sqrt{3}$
• C. $1:\sqrt{2}$
• D. $\sqrt{2}:1$

Answer 1

The Correct Option is A

(Stopping distance) $x_1 = \frac{u^2}{2 g \sin 60^{\circ}}$
(Stopping distance) $x_2 = \frac{u^2}{2 g \sin 30^{\circ}}$
$\Rightarrow \frac{x_{1}}{x_{2}} = \frac{\sin30^{\circ}}{\sin60^{\circ}} = \frac{1\times2}{2\times\sqrt{3}} = 1 : \sqrt{3} $