A particle moving horizontally with a velocity u strikes a fixed frictionless sphere at a height R/2 above the center of sphere. After striking the sphere, velocity of particle changes to vertically upward .If R be the radius of sphere, the maximum height attained by the particle with respect to center of sphere is

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$\frac{u^{2}}{6 g} + \frac{R}{2}$

$\frac{u^{2}}{2 g} + \frac{R}{2}$

$\frac{u^{2}}{4 g} + \frac{R}{2}$

$\frac{u^{2}}{3 g} + \frac{R}{2}$

answer is A.

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Detailed Solution

Since the force of contact will be only normal force,
$∴$ The velocity of the particle along the tangent will be conserved during the collision

$\begin{array}{l} Usin 30 = Vsin 60 \\ V = \frac{U}{\sqrt{3}} \end{array}$
Maximum height from centre
$= \frac{R}{2} + \frac{V^{2}}{2 g} = \frac{R}{2} + \frac{U^{2}}{6 g}$

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