A bomb at rest at the summit of a cliff breaks into two equal fragments. One of the fragments attains a horizontal velocity of 20 $\sqrt{3}m{s}^{-1}$ . The horizontal distance between the two fragments, when their displacement vectors is inclined at 60⁰ relative to each other is (g =10ms^-2 )

$$\begin{gathered} \mathrm{At} \mathrm{time} \mathrm{t} \mathrm{after} \mathrm{the} \mathrm{explosion} , \mathrm{when} \mathrm{displacement} \mathrm{vector} \mathrm{are} \mathrm{at} \mathrm{an} \mathrm{angle} \\ {60}^0 \mathrm{with} \mathrm{each} \mathrm{other} \mathrm{angle} \mathrm{with} \mathrm{the} \mathrm{vertical} \mathrm{for} \mathrm{each} \mathrm{displacement} \mathrm{vector} \\ \mathrm{is} {30}^0. \mathrm{Then} \tan {30}^0= \frac{\mathrm{ut}}{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right){\mathrm{gt}}^2}=\frac{2\mathrm{u}}{\mathrm{gt}} \\ \mathrm{so} \mathrm{t}=\frac{2\mathrm{u}\sqrt{3}}{\mathrm{g}} \\ \mathrm{seperation} \mathrm{between} \mathrm{them} =2\mathrm{ut}=\frac{2\mathrm{u}2\mathrm{u}\sqrt{3}}{\mathrm{g}}=\frac{4{\mathrm{u}}^2\sqrt{3}}{\mathrm{g}}=\frac{4\times 400\times 3\sqrt{3}}{10} \\ =480\sqrt{3} \end{gathered}$$