Two bodies are thrown simultaneously from a tower with same initial velocity v_o : one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is

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$2 v_{o} t + \frac{1}{2} {gt}^{2}$

v_ot

$v_{o} t + \frac{1}{2} {gt}^{2}$

2v_ot

answer is B.

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Detailed Solution

For vertically upward motion, $h_{1} = v_{0} t - \frac{1}{2} {gt}^{2}$ and for vertically down ward motion, $h_{2} = - v_{0} t - \frac{1}{2} {gt}^{2}$
$∴$ Total distance between the two bodies in t sec $h = h_{1} - h_{2} = 2 v_{o} t$

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