Two particles are projected simultaneously from two points O and O' such that d is the horizontal distance and h is the vertical distance between them as show in figure. These are projected at same inclination ` alpha` with the horizontal with the same speed v. Find an expression for time at which their separation becomes minimum.

Explanation

Position of particles at any time t ( taking O as origin ) is
` vecr_(1) = vt cos alpha hat I + ( vt sin alpha - 1/2 "gt"^(2)) hatj`
` vecr_(2) = ( d - vt cos alpha) hati + ( -h + vt sin alpha - 1/2 "gt"^(2)) hatj`
`vecr = vecr_(1) -vecr_(2) ( 2vt cos alpha -d) hati + h hatj`
` Rightarrow |vecr| = sqrt((2vt cos alpha -d)^(2) +h^(2)))`
Now, ` (dr)/(dt) = 0, as r = r_(min)`
` Rightarrow 1/2[(2tv cos alpha -d)^(2) +h^(2)]^(1/2 -1) xx 2 [ 2 vt cos alpha -d] xx [ 2 v cos alpha] =0`
` Rightarrow t= d/(2 v cos alpha) `