A particle is projected from point $A$ with velocity $u \sqrt{2}$ at angle $45$ to horizontal as shown in diagrame. It strikes the plane $B C$ at right angle. What is the velocity of particle at the time of impact?

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

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\frac{u}{2}

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

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Solution

The correct option is D $\frac{2 u}{\sqrt{3}}$
The horizontal component of the velocity remains the same= $\sqrt{2} u c o s 45^{\circ} = u$
Let the vertical velocity of the particle when it strikes the plane be $v$ downwards. Since it hits the plane perpendicularly, the final velocity forms an angle of $60^{\circ}$ with the vertical.
Thus $\frac{u}{v} = t a n 60^{\circ} = \sqrt{3}$
$⟹ v = \frac{u}{\sqrt{3}}$
Thus the magnitude of the velocity when it hits the plane is $\sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{u^{2} + (\frac{u}{\sqrt{3}})^{2}} = \frac{2 u}{\sqrt{3}}$

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A particle is projected from point A with velocity u√2 at angle 45 to horizontal as shown in diagrame. It strikes the plane BC at right angle. What is the velocity of particle at the time of impact?

A particle is projected from point $A$ with velocity $u \sqrt{2}$ at angle $45$ to horizontal as shown in diagrame. It strikes the plane $B C$ at right angle. What is the velocity of particle at the time of impact?