Question

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 $\theta$. The time when the distance between them is minimum is

• A. $\frac{h}{2vsin\theta }$
• B. $\frac{h}{2vcos\theta }$
• C. $\frac{h}{v}$
• D. $\frac{h}{2v}$

Answer 1

The correct option is D

$\frac{h}{2v}$

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

$s$ = $\sqrt{(h-(v-vsin\theta t){)}^2+(vcos\theta t{)}^2}$

or $s^2$ = $(h-(v-vsin\theta t){)}^2$ + $(vcos\theta t{)}^{2}s$ is minimum when

or $\frac{d}{dt}\left(s^2\right)$ = $0$

or $2(h-(v-vsin\theta \left)t\right)(vsin\theta -v)+2v^{2}cos{}^2\theta t$ = $0$

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