Question

A stone projected at an angle of ${60}^{\circ }$ from the ground level strikes a building of height $h$ at an angle of ${30}^{\circ }$. Then the speed of projection of the stone is :

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

Answer 1

The correct option is C $\sqrt{3gh}$

$\text{Step 1: Initial velocities [Ref. Fig. 1]}$

Let u be the velocity of projection, Resolving into components:

Initial velocities:

$u_x=u\cos {60}^o=\frac{u}{2}$

$u_y=u\sin {60}^{\circ }=\frac{\sqrt{3}u}{2}$

$\text{Step 2: Final velocities at the point of strike}$

Horizontal component remains constant during complete motion as there is no acceleration in horizontal direction

$v_x=u\cos \theta$ $=\frac{u}{2}$

From fig (2)

$\frac{v_y}{v_x}=\tan {30}^{\circ }$ $\text{[Ref. Fig. 2]}$

$\Rightarrow$ $v_y=\frac{u}{2\sqrt{3}}$

$\text{Step 3: Calculation of speed of projection}$

In $y$ direction, Initial velocity: $u_y=\frac{\sqrt{3}u}{2}$ Final velocity :$v_y=\frac{u}{2\sqrt{3}}$

Acceleration $a_y=-g$ $s=h$

Since acceleration is constant therefore applying equation of motion in $y$ direction

$v_y^2-u_y^2=2gs$

$\Rightarrow$ $\frac{u^2}{12}-\frac{3u^2}{4}=-2gh$

$\Rightarrow$ $u^2=3gh$

$\Rightarrow$ $u=\sqrt{3gh}$

Hence option $C$ is correct.