\[(a)\;19.4\; m \quad (b)\;9.8 m \quad (c)\;16.2 m \quad (d)\;18.6\]

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At a time when ball again meets the lift

Distance covered by lift =Distance covered by ball

$10 t+\large\frac{1}{2}$$ \times 2 \times t^2=20 t-\large\frac{1}{2}$$ \times 10 \times t^2$

We get $t=0\; or\; t= \large\frac{5}{3}$

After a time $t=5/3$ both would have been displaced by a distance

$S_L=S_b$

$\qquad=10 \times \large\frac{5}{3} +\frac{1}{2} $$\times 2 \times $$ \bigg(\large\frac{5}{3}\bigg)^2$

$\qquad=\large\frac{175}{9}$

$\qquad=19.4 m$

Hence a is the correct answer.

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