Question

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 :

• A. $\frac{\sqrt{3} u }{2}$
• B. $\frac{ u }{2}$
• C. $\frac{2 u}{\sqrt{3}}$
• D. $u$

Answer 1

Answer: (C)

Let $v$ be the velocity at the time of collision.
Then, $u\sqrt{2}$ cos $45^{\circ} = v$ sin $60^{\circ}$

$( u \sqrt{2})\left(\frac{1}{\sqrt{2}}\right)=\frac{\sqrt{3} v }{2}$
$\therefore v=\frac{2}{\sqrt{3}} u$