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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 projectile is thrown with speed $40 \,ms ^{-1}$ at angle $\theta$ from horizontal. It is found that projectile is at same height at $1 \,s$ and $3 \,s$. What is the angle of projection?
  • A
    $\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
  • ✓
    $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
  • C
    $\tan ^{-1}(\sqrt{3})$
  • D
    $\tan ^{-1}(\sqrt{2})$
✓

Answer

Correct option: B.
$\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
b
(b)

$\tan \theta=\frac{v_y}{v_x}$

Also, $t_1+t_2=\frac{2 u \sin \theta}{g}$

$4=\frac{2 \times 40 \times \sin \theta}{10}$

$\sin \theta=\frac{1}{2} \Rightarrow \theta=30^{\circ}$

So, $\tan \theta=\tan 30^{\circ} \Rightarrow \frac{1}{\sqrt{3}}$

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

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