Question

When a body is shot from the bottom of a long smooth inclined plane kept at an angle $30^{\circ}$ with horizontal, it can travel a distance $x$ along the plane. But when the inclination is increased to $60^{\circ}$ and the same body is shot with the same velocity, it can travel $y$ distance then $x: y$ will be

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

Answer 1

Answer: (A)

$\frac{x}{y}=\frac{u^2}{2 g \sin 30^{\circ}}=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}$
$=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\sqrt{3}}{2} \times \frac{2}{1}=\sqrt{3}$
$\frac{u^2}{2 g \sin 60^{\circ}}$