Question

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. $\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
• B. $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
• C. $\tan ^{-1}(\sqrt{3})$
• D. $\tan ^{-1}(\sqrt{2})$

Answer 1

Answer: (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, $\tan \theta=\tan 30^{\circ}$
$\Rightarrow \frac{1}{\sqrt{3}}$
$\theta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$