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