A particle is projected vertically upwards from O with velocity v… - Vidyadip

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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