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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 :
292984_8302add190664bafaa97a80e27e9a073.png


A
2gh
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B
6gh
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C
3gh
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D
gh
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Solution

The correct option is C $$\sqrt{3gh}$$
$$\textbf{Step 1: Initial velocities [Ref. Fig. 1]}$$
Let u be the velocity of projection, Resolving into components:
Initial velocities:
$$u_x = u \cos 60^o=\dfrac{u}{2}$$
$$u_y = u \sin 60^{\circ}= \dfrac{\sqrt{3}u}{2}$$

$$\textbf{Step 2: Final velocities at the point of strike}$$
Horizontal component remains constant during complete motion as there is no acceleration in horizontal direction
 $$v_x = u \cos \theta$$   $$ = \dfrac{u}{2}$$
From fig (2)
             $$\dfrac{v_y}{v_x} = \tan 30^{\circ}$$         $$\textbf{ [Ref. Fig. 2]}$$

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

$$\textbf{Step 3: Calculation of speed of projection}$$
In $$y$$ direction, Initial  velocity: $$u_y = \dfrac{\sqrt{3} u}{2}$$                Final velocity :$$v_y = \dfrac{u}{2\sqrt{3}}$$
Acceleration $$a_y = -g$$        $$s = h$$
Since acceleration is constant therefore applying equation of motion in $$y$$ direction
             $$v_y^2 - u_y^2 = 2gs$$

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

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

          $$\Rightarrow$$   $$u = \sqrt{3 gh}$$
Hence option $$C$$ is correct.

2107771_292984_ans_f6f4bef892d0486a967b2aa89efdab1b.png

Physics

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