Two bodies are thrown simultaneously from a tower with same initial velocity $ν_{0} :$ one vertically upwards, the other vertically downwards. The distance between the two bodies after t is

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$2 ν_{0} t + \frac{1}{2} g t^{2}$

$2 ν_{0} t$

$ν_{0} t + \frac{1}{2} g t^{2}$

$ν_{0} t$

answer is B.

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Detailed Solution

For vertically upward motion,
$h_{1} = ν_{0} t - \frac{1}{2} {gt}^{2}$ and
For vertically downward motion,
$h_{2} = ν_{0} t + \frac{1}{2} {gt}^{2}$
$∴$ Total distance covered in t sec

$h = h_{1} + h_{2} = 2 ν_{0} t .$

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