The vertical distance between the masses is $\frac{20}{3}$m. The time at which the distance becomes zero is

For 1 kg mass and 2 kg mass, the acceleration of the system is given by the formula

$\mathrm{a}=\frac{\mathrm{g}\left({\mathrm{m}}_1-{\mathrm{m}}_2\right)}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}$

$\mathrm{a}=\frac{10\times \left(2-1\right)}{\left(2+1\right)}=\frac{10}{3}{\mathrm{ms}}^{-2}$

Acceleration of 1 kg mass $=\frac{10}{3}{\mathrm{ms}}^{-2 }$up $=\frac{-10}{3}{\mathrm{ms}}^{-2 }$

Acceleration of 2 kg mass$=\frac{10}{3}{\mathrm{ms}}^{-2 }$ down $=\frac{+10}{3}{\mathrm{ms}}^{-2 }$

$\therefore$Relative acceleration of 2 kg mass w. r. t 1 kg mass$=\frac{+10}{3}-\left(-\frac{10}{3}\right)=\frac{10}{3}+\frac{10}{3}=\frac{20}{3}{\mathrm{ms}}^{-2 }$

Now :

$\mathrm{U}=0; \mathrm{a}=\frac{20}{3}{\mathrm{ms}}^{-2}; \mathrm{s}=20\mathrm{m};\mathrm{t}=?$

$\mathrm{S}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

$\therefore \frac{20}{3}=0\times \mathrm{t}+\frac{1}{2}\times \frac{20}{3}\times {\mathrm{t}}^2$

$\therefore \frac{20}{3}=\frac{10}{3}\times {\mathrm{t}}^2$

$\therefore 2={\mathrm{t}}^2$

$\therefore \mathrm{t}=\sqrt{2}\mathrm{s}$