A ball dropped from a building of height h bounces with the same speed from a surface inclined at angle $α$ and then hits a vertical wall at a distance d from the point P, as shown in the figure.
If the velocity of the ball when it hits the wall is horizontal, then:

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$α = \frac{1}{4} \sin^{- 1} (\frac{d}{h})$

$α = \frac{1}{2} \sin^{- 1} (\frac{d}{h})$

$α = \frac{1}{2} \sin^{- 1} (\frac{2 d}{h})$

$α = \frac{1}{2} \sin^{- 1} (\frac{4 d}{h})$

answer is A.

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Detailed Solution

The ball strikes the surface with speed $v = \sqrt{2 g h}$ . at angle a with the normal to the surface. The ball then rebounds at the same speed at angle $2 α$ with the vertical; that is, at angle $θ = 90^{\circ} - 2 a$ with the horizontal. Now the ball moves in a parabolic trajectory of range 2d.Hence

$\begin{array}{ll} \frac{v^{2} \sin (2 θ)}{g} = 2 d \\ \Rightarrow \frac{2 g h \sin (180^{\circ} - 4 α)}{g} = 2 d \\ \Rightarrow \sin (4 α) = \frac{d}{h} \\ \Rightarrow α = \frac{1}{4} \sin^{- 1} (\frac{d}{h}) \end{array}$