A particle is projected up on a inclined plane. The component of its velocity perpendicular to the plane when it strikes the plane is 10m/s. If the particle is projected at an angle of $30^{0},$ find velocity of projection (m/s) :

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$10 \sqrt{2}$

$20 \sqrt{2}$

answer is C.

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Detailed Solution

The component of velocity perpendicular to inclined plane at the point of projection and the point of collision with the plane remains same, i.e, $v \sin θ$

Given

$\begin{array}{l} v \sin θ = 10 \\ v = \frac{10}{\sin θ} = \frac{10}{\sin 30^{0}} = 20 m / \sec \end{array}$

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