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
h2vsinθ
b
h2vcosθ
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−vsinθt}2+(vcosθt)2or s2={h−(v−vsinθ)t}2+(vcosθt)2s is minimum when ddts2=0 or 2{h−(v−vsinθ)t}{vsinθ−v}+2v2cos2θt=0 or t=h2v