$\therefore$ Acceleration of aeroplane with respect to the earth,

$a_{A\to E}=5 {\text{m s}}^{-2}$

Acceleration of stone with respect to the earth,

$a_{S{\to E}_{ }}=-10 {\text{m s}}^{-2}$

or $a_{E{\to S}_{ }}=-a_{S{\to E}_{ }}=10 {\text{m s}}^{-2}$

Thus, the acceleration of aeroplane with respect to the stone,

$a_{A{\to S}_{ }}=a_{A\to E}+a_{E\to S}=5+10=15 {\text{m s}}^{-2}$

$\therefore$ The relative separation between the stone and aeroplane in $\mathrm{2 }\text{s}$,

$s=\frac{1}{2}\left(a_{A\to S}\right)t^2=\frac{1}{2}\times 15\times {\left(2\right)}^2=30 \text{m}$