Two particles are projected simultaneously from two points O and O' such that 20 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 600 to the horizontal with the same velocity 10 ms-1. Find the time (in s) after which their separation becomes minimum.

Let us find relative velocity of O w.r.t. O'.v→O/O'=v→O−v→O'=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 = 20 mTime take=20vO/O'=2010=2s