particle is projected vertically upwards from O with velocity v and a second particle projected at same instant from P at height h above O with velocity v and angle of projection theta . find tme after which distance between the 2 s minimum ?

position vector of 2nd particle= vtcos(theta)i +(h+(vtsinθ -(g/2)t^2)j,position vector of first particle=vt-(g/2)t^2 j

subtract one position vector from the other and obtain the relative positive vector as a function of t, minimize it with respect to find the answer, vector connecting particles= vtcos(theta)i +(h-vt(1-sin(theta))