A particle is projected from a point $A$ with velocity $u\sqrt{2}$ at an angle of $45{^\circ}$ with horizontal as shown in figure. It strikes the plane $BC$ at right angles. The velocity of the particle at the time of collision is [image:4_image27.emf]

Question

Dr. Abubacker · Accepted Answer

(c) Let $v$ be the velocity at the time of collision
[image:4_image120.emf]
Then, $u\sqrt{2}\cos{45{^\circ}} = v\sin{60{^\circ}}$
$\left( u\sqrt{2} ight)\left( \frac{1}{\sqrt{2}} ight) = \frac{\sqrt{3}v}{2}\ \ \ 	herefore v	ext{=}\frac{2}{\sqrt{3}}u\ $

Q. A particle is projected from a point $A$ with velocity $u\sqrt{2}$ at an angle of $45{^\circ}$ with horizontal as shown in figure. It strikes the plane $BC$ at right angles. The velocity of the particle at the time of collision is

Explanation

Reviewed by Dr. Abubacker

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