Two particles are projected simultaneously from two points O and O' such that 10 m is the horizontal and 5 m is the vertical distance between them as shown in the figure. They are projected at the same inclination 60^o to the horizontal with the same velocity 10 ms^-1. Find the time after which their separation becomes minimum.

Let us find relative velocity of O. w.r.t. O'.

$\begin{array}{l}{\overrightarrow{v}}_{\text{olo }}={\overrightarrow{v}}_o-\overrightarrow{v_o}=\left(10\cos {60}^{\circ }\widehat{i}+10\sin {60}^{\circ }\widehat{j}\right)\\ -\left[10\cos {60}^{\circ }(-\widehat{i})+10\sin {60}^{\circ }\widehat{j}\right]=10\widehat{i}\end{array}$

This is along horizontal direction.
If we assume O' to be at rest, then O will move along OA w.r.t. O' and minimum separation will be when particle O is at A.

$\text{ For this, relative displacement travelled }=OA=10\mathrm{m}$

$\text{ Time taken }=\frac{10}{v_{O/O^{\mathrm{\prime }}}}=\frac{10}{10}=1\mathrm{s}$