A body falling under gravity covers two points A and B separated by...

From $A \to B$, using the equation of motion, we have

$- 80 = - {v}_{1}t - \dfrac{1}{2} \times 10{t}^{2} ⇒ - 80 = - 2{v}_{1} - \dfrac{1}{2} \times 10 \times {2}^{2} ⇒ - 80 = - 2{v}_{1} - 20 ⇒ - 60 = - 2{v}_{1} ⇒ {v}_{1} = 30m{s}^{- 1}$

From $O$ to $A$, it follows that

${v}^{2} = {u}^{2} + 2gS ⇒ {30}^{2} = 0 + 2 \times ( - 10 ) ( - S ) ⇒ 900 = 20S ⇒ S = 45m$