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3-2.Motion in Plane
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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
$h/v$
D
$h/2v$
Solution
In from of ball thrown from $p$
$v_{\text {resultant }}=2 v \cos \left(\frac{90^{\circ}+\theta}{2}\right)$
$=2 v \cos \left(45^{\circ}+\theta / 2\right)$
$t=\frac{d}{v_{\text {resultant }}}=\frac{h \cos \left(45^{\circ}+\frac{\theta}{2}\right)}{2 v \cos \left(45^{\circ}+\frac{\theta}{2}\right)}=\frac{h}{2 v}$
Standard 11
Physics
Similar Questions
Match the columns
| Column $-I$ $R/H_{max}$ |
Column $-II$ Angle of projection $\theta $ |
| $A.$ $1$ | $1.$ ${60^o}$ |
| $B.$ $4$ | $2.$ ${30^o}$ |
| $C.$ $4\sqrt 3$ | $3.$ ${45^o}$ |
| $D.$ $\frac {4}{\sqrt 3}$ | $4.$ $tan^{-1}\,4\,=\,{76^o}$ |
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