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A particle projected from ground moves at angle $45^{\circ}$ with horizontal one second after projection and speed is minimum two seconds after the projection. The angle of projection of particle is [Neglect the effect of air resistance]
$\tan ^{-1}(3)$
$\tan ^{-1}(2)$
$\tan ^{-1}(\sqrt{2})$
$\tan ^{-1}(4)$
Solution
(b)
$\theta=45^{\circ}, t=1 \,s$
$\tan \phi=\frac{V_y}{U_y}=\frac{u \sin \theta-g t}{u \cos \theta}$
$\tan 45^{\circ}=\frac{u \sin \theta-g \times 1}{u \cos \theta} \Rightarrow u \cos \theta=u \sin \theta-g$
also, $V_y=0$, after $1^{\text {st }}$ (as speed is minimum)
$u \sin \theta-g \times 2=0$
$\Rightarrow u \sin \theta=2 g \ldots (i)$
so, $u \cos \theta=2 g-g$
$u \cos \theta=g \ldots (ii)$
so, $\frac{(i)}{(ii)}=\frac{u \sin \theta}{u \cos \theta}=\frac{2 g}{g}$
$\Rightarrow \tan \theta=2$
$\theta=\tan ^{-1}(2)$
