Question
In the figure shown, the two projectiles are fired simultaneously. The minimum distance between them during their flight is ........ $m$
✓
Answer
Talking origin at a and $x$ axis along $A B$ Velocity of $A$ $w.r.t.$ $B$ $=20 \sqrt{3}(\cos 60 \hat{i}+\sin 60 \hat{j})-20(\cos 150 \hat{i}+\sin 150 \hat{j})$
$=20 \sqrt{3}\left(\frac{1}{2} \hat{i}+\frac{\sqrt{3}}{2} \hat{j}\right)-20\left(-\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}\right)=20 \sqrt{3} \hat{i}+20 \hat{j}$
$\tan \theta=\frac{20}{20 \sqrt{3}}=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ}$
so $\frac{d_{\min }}{20}=\sin \theta=\sin 30^{\circ}=\frac{1}{2} \Rightarrow d_{\min }=10 \mathrm{m}$
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