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:

Accepted Answer

$$ ext{(Stopping distance)} \quad x_1 = \frac{u^2}{2g\sin 60^\circ}$$

$$ ext{(Stopping distance)} \quad x_2 = \frac{u^2}{2g\sin 30^\circ}$$

$$\Rightarrow \frac{x_1}{x_2} = \frac{\sin 30^\circ}{\sin 60^\circ} = \frac{1 imes 2}{2 imes \sqrt{3}} = 1 : \sqrt{3}$$