AB is an inclined plane of inclination 30^o with horizontal. Point O is 20 m above point A. A particle is projected horizontally from O leftwards, and it collides with the plane AB perpendicularly. Speed of the particle at the time of projection should be $(g = 10 m / s^{2})$

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$13 m / s$

$8 \sqrt{3} m / s$

$4 \sqrt{5} m / s$

$2 \sqrt{3} m / s$

answer is C.

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Detailed Solution

$In x direction: v_{I} = u_{I} + a_{I} t$

$\Rightarrow 0 = u \cos 30^{\circ} - g \sin 30^{\circ} t \Rightarrow u \sqrt{3} = g t$ ……..(i)

$\begin{array}{l} In y direction: s_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2} \\ \Rightarrow 20 \cos 30^{\circ} = u \cos 60^{\circ} t + \frac{1}{2} g \cos 30^{0} t^{2} \\ \Rightarrow 20 \sqrt{3} = u t + \frac{1}{2} g t^{2} \sqrt{3} \\ \Rightarrow 20 \sqrt{3} = \frac{u^{2} \sqrt{3}}{g} + \frac{\sqrt{3}}{2} g \cdot \frac{3 u^{2}}{g^{2}} \Rightarrow 20 = \frac{5 u^{2}}{2 g} \\ \Rightarrow u = \sqrt{80} m / s = 4 \sqrt{5} m / s \end{array}$