Question

A stone projected at an angle of 60° from the ground level strikes at an angle of 30° on the roof of a building of height 'h'. Then the speed of projection of the stone is:

• A. $\sqrt{2gh}$
• B. $\sqrt{6gh}$
• C. $\sqrt{3gh}$
• D. $\sqrt{gh}$

Answer 1

Answer: (C)

$\sqrt{3gh}$

Answer 2

Let the initial speed of projection be $v_0$ and the angle of projection be $\theta = 60^\circ$. The horizontal component of initial velocity is $v_{0x} = v_0 \cos 60^\circ = \frac{v_0}{2}$, and the vertical component is $v_{0y} = v_0 \sin 60^\circ = \frac{\sqrt{3}v_0}{2}$.

At height $h$, the horizontal component of velocity remains constant: $v_x = v_{0x} = \frac{v_0}{2}$. The vertical component of velocity at height $h$, $v_y$, can be found using $v_y^2 = v_{0y}^2 - 2gh$. $v_y^2 = \left(\frac{\sqrt{3}v_0}{2}\right)^2 - 2gh = \frac{3v_0^2}{4} - 2gh$.

The stone strikes the roof at an angle of $30^\circ$ with the horizontal. Therefore, $\tan(30^\circ) = \frac{v_y}{v_x}$. $v_y = v_x \tan(30^\circ) = \left(\frac{v_0}{2}\right) \left(\frac{1}{\sqrt{3}}\right) = \frac{v_0}{2\sqrt{3}}$.

Squaring $v_y$: $v_y^2 = \left(\frac{v_0}{2\sqrt{3}}\right)^2 = \frac{v_0^2}{12}$.

Equating the two expressions for $v_y^2$: $\frac{v_0^2}{12} = \frac{3v_0^2}{4} - 2gh$ $2gh = \frac{3v_0^2}{4} - \frac{v_0^2}{12} = \frac{9v_0^2 - v_0^2}{12} = \frac{8v_0^2}{12} = \frac{2v_0^2}{3}$.

Solving for $v_0^2$: $v_0^2 = \frac{3}{2} \times 2gh = 3gh$. Therefore, the speed of projection $v_0 = \sqrt{3gh}$.