Let the body hit the floor.

The distance travelled by the body in $t$ seconds is given by $d(t)=\frac12 gt^2$ . Thus, the distance travelled during the $t^\textrm{th}$ second is $\begin{aligned} \Delta d(t) &= d(t) - d(t-1) \\ &= \frac12 g\left(t^2 - \left(t-1\right)^2\right) \\ &= \frac12 g \left(2t-1\right) \end{aligned}$ The ratio of distance travelled in the $t^\textrm{th}$ second to that travelled by the $t^\textrm{th}$ second is $\frac{d(t)}{\Delta d(t)} = \frac{t^2}{2t-1}$ This ratio is equal to 4/3 when $t=2$ . Thus, the distance travelled in $n = t+3$ seconds is $d(5)=125$ m.