A small ball thrown with an initial velocity u directed at an angle $θ$ = 37° above the horizontal collides inelastically (e = 1/4) with a vertical massive wall moving with a uniform horizontal velocity u/5 towards ball. After collision with the wall, the ball returns to the point from where it was thrown. Neglect friction between ball and wall. The time t from beginning of motion of the ball till the moment of its impact with the wall is (tan37° = 3/4)

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$\frac{18 u}{25 g}$

$\frac{54 u}{145 g}$

$\frac{54 u}{25 g}$

$\frac{3 u}{5 g}$

answer is C.

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Detailed Solution

Since the wall is frictionless , time of flight T = $\frac{2 u \sin 37^{0}}{g} = \frac{6 u}{5 g}$

Horizontal component of velocity of the ball relative to the wall just after the impact =e $u_{r e l} = \frac{1}{4} (\frac{u}{5} + u \cos 37^{0}) = \frac{u}{4}$

Therefore after the impact ,horizontal component of velocity of the ball relative to the ground = $\frac{u}{5} + \frac{u}{4} = \frac{9 u}{20}$

So,( u cos $37^{0} + \frac{u}{5}) t = \frac{9 u}{20} (\frac{6 u}{5 g} - t) \Rightarrow t = \frac{54 u}{145 g}$

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