A particle is projected vertically upwards from O with velocity v and a second p

Question

A particle is projected vertically upwards from

O

with velocity

v

and a second particle is projected at the same instant from

P

(at a height h above

O

) with velocity

v

at an angle of projection

θ

. The time when the distance between them is minimum is

Options:

A . h2v sinθ

B . h2v cosθ

C . hv

D . h2v

Answer: Option D
:
D
Relative acceleration between the two particles is zero. The distance between then at time

t

s

\sqrt{(h - (v - v s i n θ t))^{2} + (v c o s θ t)^{2}}

s^{2}

(h - (v - v s i n θ t))^{2}

(v c o s θ t)^{2} s

is minimum when
or

\frac{d}{d t} (s^{2})

0

2 (h - (v - v s i n θ) t) (v s i n θ - v) + 2 v^{2} c o s^{2} θ t

0

t

\frac{h}{2 v}

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