In the figure shown,the two projectiles are f | vedclass

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(C) Let the origin be at the left projectile $A$ and the $x$-axis be along the line connecting $A$ and $B$. The initial position of $A$ is $(0, 0)$ an

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$10$

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(C) Let the origin be at the left projectile $A$ and the $x$-axis be along the line connecting $A$ and $B$. The initial position of $A$ is $(0, 0)$ and $B$ is $(20, 0)$.The velocity of $A$ is $\vec{v}_A = 20\sqrt{3}(\cos 60^{\circ} \hat{i} + \sin 60^{\circ} \hat{j}) = 20\sqrt{3}(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}) = 10\sqrt{3} \hat{i} + 30 \hat{j}$.The velocity of $B$ is $\vec{v}_B = 20(\cos 150^{\circ} \hat{i} + \sin 150^{\circ} \hat{j}) = 20(-\frac{\sqrt{3}}{2} \hat{i} + \frac{1}{2} \hat{j}) = -10\sqrt{3} \hat{i} + 10 \hat{j}$.The relative velocity of $A$ with respect to $B$ is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B = (10\sqrt{3} - (-10\sqrt{3})) \hat{i} + (30 - 10) \hat{j} = 20\sqrt{3} \hat{i} + 20 \hat{j}$.The relative initial position is $\vec{r}_{AB} = \vec{r}_A - \vec{r}_B = (0 - 20) \hat{i} + (0 - 0) \hat{j} = -20 \hat{i}$.The minimum distance is the perpendicular distance from the origin to the line of relative motion,given by $d_{\min} = \frac{|\vec{r}_{AB} 	imes \vec{v}_{AB}|}{|\vec{v}_{AB}|}$.$|\vec{r}_{AB} 	imes \vec{v}_{AB}| = |(-20 \hat{i}) 	imes (20\sqrt{3} \hat{i} + 20 \hat{j})| = |-400 \hat{k}| = 400$.$|\vec{v}_{AB}| = \sqrt{(20\sqrt{3})^2 + 20^2} = \sqrt{1200 + 400} = \sqrt{1600} = 40$.$d_{\min} = \frac{400}{40} = 10 \ m$.

In the figure shown,the two projectiles are fired simultaneously. The minimum distance between them during their flight is ........ $m$.

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