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(C) Let $v$ be the velocity of the particle at the time of collision.Since the particle strikes the plane $BC$ at right angles,the component of veloci

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$2u$

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(C) Let $v$ be the velocity of the particle at the time of collision.Since the particle strikes the plane $BC$ at right angles,the component of velocity parallel to the plane $BC$ must be zero at the moment of impact.The plane $BC$ is inclined at an angle of $60^{\circ}$ with the horizontal.The horizontal component of the initial velocity is $u_x = u\sqrt{2} \cos 45^{\circ} = u\sqrt{2} \cdot \frac{1}{\sqrt{2}} = u$.At the point of collision,the velocity vector $v$ makes an angle of $60^{\circ}$ with the vertical (since it is perpendicular to the plane inclined at $60^{\circ}$ to the horizontal).Therefore,the horizontal component of the final velocity is $v_x = v \sin 60^{\circ}$.Since there is no acceleration in the horizontal direction,the horizontal component of velocity remains constant throughout the motion.Thus,$v_x = u_x$.$v \sin 60^{\circ} = u$$v \cdot \frac{\sqrt{3}}{2} = u$$v = \frac{2u}{\sqrt{3}}$Wait,re-evaluating the geometry: If the plane is inclined at $60^{\circ}$ to the horizontal,the normal to the plane makes an angle of $30^{\circ}$ with the vertical or $60^{\circ}$ with the horizontal. If the velocity $v$ is perpendicular to the plane,it makes an angle of $60^{\circ}$ with the horizontal. Thus,the horizontal component is $v \cos 60^{\circ} = u$. Therefore,$v \cdot (1/2) = u$,which gives $v = 2u$.

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

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