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A ball is thrown up velocity 20 m/s relative to ground in open lift moving up with speed 10 m/s at time of throwing and has an acceleration of 2m/s² upward, ball meets
lift after

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$\frac{3}{5} s$

$\frac{5}{3} s$

$\frac{2}{5} s$

$\frac{5}{2} s$

answer is C.

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Detailed Solution

$T = \frac{24_{rel}}{arel} = \frac{2 \times 10}{12} = \frac{5}{3} s$

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