A man in a lift ascending with an upward acceleration o throws a ball vertically upwards with a velocity v and catches it after $t_{1}$ seconds. Afterwards when the lift is descending with the same acceleration a acting downwards the man again throws the ball vertically unward with the same velocity and catches it after t₂ second.

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the acceleration of the ball is g $(\frac{t_{1}}{t_{2}} + \frac{t_{2}}{t_{1}})$

the acceleration of the ball is $g (t_{2} - t_{1}) / (t_{2} + t_{1})$

the velocity of the ball is ${gt}_{1} t_{2} / (t_{1} + t_{2})$

the velocity of the ball is $g (t_{1} + t_{2}) / t_{1} t_{2}$

answer is C.

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Detailed Solution

When the lift is ascending with acceleration a, then time of flight of vertical motion
$t_{1} = \frac{2 v}{(g + a)} (∵ time of flight = \frac{2 v}{g})$
When the lift is descending with acceleration o
$t_{2} = \frac{2 v}{(g - a)}$

Solving these equation, we get
$V = \frac{{gt}_{1} t_{2}}{t_{1} + t_{2}}$ and $a = \frac{g (t_{2} - t_{1})}{(t_{1} + t_{2})}$