A particle 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 particle 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)/(2upsilonsintheta)`
B. `(h)/(2upsiloncostheta)`
C. `h//upsilon`
D. `(h)/(2upsilon)`

Answer

D

Explanation

Relative acceleration between the particles is zero. The distance between them at time t is
`s =sqrt({h-(upsilon-upsilonsintheta)t}^(2)+(upsiloncosthetat)^(2))" or "`
`s^(2)={h-(upsilon-upsilonsintheta)t}^(2)+(upsiloncosthetat)^(2)s`
`(ds^(2))/(dt)=0`
`2{h-(upsilon-upsilonsintheta)t}(upsilonsintheta-upsilon)+2upsilon^(2)cos^(2)thetat=0t=(h)/(2upsilon)`