A particlele is projected vertically upwards from the ground with a speed V and a speed v and a second particle is projected at the same instant from a height h directly above the first particel with the same speed v at an angle of projection `theta` with the horizontal in upwards direction. The time when the distance between them is minimum is

Options

A. `(h)/(2v sin theta)`
B. `(h)/(2v cos theta)`
C. `(h)/(sqrt2 v)`
D. `(h)/(2 v)`

Answer

D

Explanation

In the relative frame
`vec(S)_(0)=-hhat(j), vec(u) _(ret) = v cos theta hat(i)+(vsin theta-v)hat(j), vec(a)_(ret) = 0`