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Physics-

General

Easy

Question

A particle is projected from a point $A$ with velocity $u square root of 2$ at an angle of $45 degree$ with horizontal as shown in figure. It strikes the plane $B C$ at right angles. The velocity of the particle at the time of collision is

$\frac{\sqrt{3 u}}{2}$
$\frac{u}{2}$
$\frac{2 u}{\sqrt{3}}$
$u$

The correct answer is: $fraction numerator 2 u over denominator square root of 3 end fraction$

Let $v$ be the velocity at the time of collision

Then, $u square root of 2 cos invisible function application 45 degree equals v sin invisible function application 60 degree$
$open parentheses u square root of 2 close parentheses open parentheses fraction numerator 1 over denominator square root of 2 end fraction close parentheses equals fraction numerator square root of 3 v over denominator 2 end fraction blank therefore v equals fraction numerator 2 over denominator square root of 3 end fraction u$

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