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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Solution

The correct option is D

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

$s$ = $\sqrt{(h - (v - v s i n θ t))^{2} + (v c o s θ t)^{2}}$

or $s^{2}$ = $(h - (v - v s i n θ t))^{2}$ + $(v c o s θ t)^{2} s$ is minimum when

or $\frac{d}{d t} (s^{2})$ = $0$

or $2 (h - (v - v s i n θ) t) (v s i n θ - v) + 2 v^{2} c o s^{2} θ t$ = $0$

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

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