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
2gh
B
6gh
C
3gh
D
gh

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.Physics

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