Q. 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:
Solution:
$($ 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{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1 \times 2}{2 \times \sqrt{3}}=1: \sqrt{3}
$
