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 $1:\sqrt{3}$

Using the third equation of motion we get,

$v^2=u^2-2as$

Acceleration along the plane is $a=gsin\theta$

Therefore, $s=\frac{u^2}{2gsin\theta }$

Since $v=0$

So,

$\frac{x_1}{x_2}=\frac{sin{\theta }_2}{sin{\theta }_1}=\frac{sin{30}^0}{sin{60}^0}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}$.