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

Gujarat BoardEnglish MediumSTD 11 SciencePhysics3.2 , motion in plane1 Mark
MCQ
A projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall at its highest point as shown in the figure.The angle $\theta$ at which the projectile is thrown is given by
  • A
    $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
  • B
    $\tan ^{-1}(\sqrt{3})$
  • ✓
    $\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
  • D
    $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
✓

Answer

Correct option: C.
$\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
c
(c)

$\frac{R / 2}{H}=\frac{\sqrt{3} H}{H}=\sqrt{3}$

or $\frac{\left(v_0^2 \sin \theta \cos \theta\right) / g}{\left(v_0^2 \sin ^2 \theta\right) / 2 g}=\sqrt{3}$

$2 \cot \theta=\sqrt{3}$

or $\tan \theta=\frac{2}{\sqrt{3}}$

or $\theta=\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$

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