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 60o 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'.v→olo =v→o−vo→=10cos⁡60∘i^+10sin⁡60∘j^−10cos⁡60∘(−i^)+10sin⁡60∘j^=10i^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. For this, relative displacement travelled =OA=10m Time taken =10vO/O′=1010=1s