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Q.

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

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a

h2vsin⁡θ

b

h2vcos⁡θ

c

h/v

d

h/2v

answer is D.

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Detailed Solution

Relative acceleration between the two particles is zero. The distance between then at time t iss=h−(v−vsin⁡θt}2+(vcos⁡θt)2or s2={h−(v−vsin⁡θ)t}2+(vcos⁡θt)2s is minimum when ddts2=0 or  2{h−(v−vsin⁡θ)t}{vsin⁡θ−v}+2v2cos2⁡θt=0 or t=h2v
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