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

The velocities of the two projectiles are:  ${\mathrm{v}}_{\mathrm{A}}=20\sqrt{3}{ \mathrm{ms}}^{-1} \text{and }{\mathrm{v}}_{\mathrm{B}}=20{ \mathrm{ms}}^{-1}$.

Horizontal components of the velocities are: ${\mathrm{v}}_{\mathrm{Ax}}=10\sqrt{3}{ \mathrm{ms}}^{-1} \text{and }{\mathrm{v}}_{\mathrm{Bx}}=-10\sqrt{3}{ \mathrm{ms}}^{-1}$.

Similarly, vertical components of the velocities are:  $v_{Ay}=30 {\mathrm{ms}}^{-1} \text{and }{v}_{By}=10 {\mathrm{ms}}^{-1}$.

The distance between the two projectiles at the time of firing is $\mathrm{d}=20\sqrt{3} \text{m}$. 

Angle made by relative velocity $v_{AB}$ with horizontal:  $\theta ={\tan }^{-1}\left(\frac{v_{Ay}-v_{By}}{v_{Ax}-v_{Bx}}\right)={\tan }^{-1}\left(\frac{30-10}{10\sqrt{3}+10\sqrt{3}}\right)={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)=30°$.

Minimum distance between A and B is given by  ${\mathrm{d}}_{\min }=d\sin \theta$$=20\sqrt{3}\times \frac{1}{2}=10\sqrt{3} \mathrm{m}$.