A particle is projected vertically upwards from O with velocity v and a second particle is projected at the same instant from P (at a height h above O) with velocity v at an angle of projection $θ$ . The time when the distance between them is minimum is

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$\frac{h}{2 v \sin θ}$

$\frac{h}{2 v \cos θ}$

$h / v$

$h / 2 v$

answer is D.

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Detailed Solution

Relative acceleration between the two particles is zero. The distance between then at time t is

$s = \sqrt{\{h - (v - v \sin θ t}^{2} + (v \cos θ t)^{2}}$

$or s^{2} = {h - (v - v \sin θ) t}^{2} + (v \cos θ t)^{2}$

$s is minimum when \frac{d}{d t} (s^{2}) = 0$

$or 2 {h - (v - v \sin θ) t} {v \sin θ - v} + 2 v^{2} \cos^{2} θ t = 0$

$or t = \frac{h}{2 v}$

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