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3.2 , motion in plane — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysics3.2 , motion in plane1 Mark
MCQ
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]
  • A
    $\tan ^{-1}(3)$
  • ✓
    $\tan ^{-1}(2)$
  • C
    $\tan ^{-1}(\sqrt{2})$
  • D
    $\tan ^{-1}(4)$
✓

Answer

Correct option: B.
$\tan ^{-1}(2)$
b
(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, $\frac{(i)}{(ii)}=\frac{u \sin \theta}{u \cos \theta}=\frac{2 g}{g}$

$\Rightarrow \tan \theta=2$

$\theta=\tan ^{-1}(2)$

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