MCQ
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 $\theta$ . The time when the distance between them is minimum is
  • A
    $\frac{h}{{2v\sin \,\theta }}$
  • B
    $\frac{h}{{2v\cos \,\theta }}$
  • C
    $h/v$
  • $h/2v$

Answer

Correct option: D.
$h/2v$
d
In from of ball thrown from $p$

$v_{\text {resultant }}=2 v \cos \left(\frac{90^{\circ}+\theta}{2}\right)$

$=2 v \cos \left(45^{\circ}+\theta / 2\right)$

$t=\frac{d}{v_{\text {resultant }}}=\frac{h \cos \left(45^{\circ}+\frac{\theta}{2}\right)}{2 v \cos \left(45^{\circ}+\frac{\theta}{2}\right)}=\frac{h}{2 v}$

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