Question

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 is 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. $1:2\sqrt{3}$

Answer 1

The correct option is C $1:\sqrt{3}$

$v^2=u^2-2as\Rightarrow s=\frac{u^2}{2a}=\frac{u^2}{2g\sin \theta }\frac{x_1}{x_2}=\frac{\sin {\theta }_2}{\sin {\theta }_1}=\frac{\sin 30°}{\sin 60°}=\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}$