# Relative amount of time spent during free fall

I am trying to understand the relative time spent during free fall.

Suppose I drop an object at height $h$ (constant gravity). Then relative amount of time the object is in free fall for the last third of the distance is given by the following, where $t$ denotes total time and $t'$ denotes the time in the last third.

$$h=\frac{1}{2}gt^2$$

$$\frac{2}{3}h=\frac{1}{2}g(t-t')^2$$

Thus,

$$\frac{t'}{t}=1\pm\sqrt{\frac{2}{3}}\approx\{0.2,1.8\}$$

That is, only $20\%$ of the time is spent in the last third, which makes sense, as the object has gained velocity in the first two thirds.

What does the other solution, $\frac{t'}{t}\approx 1.8$, signify?

Given your definitions of $t$ and $t'$, the result $\frac{t'}{t}\approx1.8$ would mean that the object spent more time in the last $1/3$ of the freefall than in the total freefall... that's clearly not possible, so what you have is simply an unphysical mathematical consequence of the quadratic formula.
Another way to set this up so you don't run into this issue would be to redefine $t'$ to be the time it takes to fall the first $2/3$, rearrange your first formula for each $t$ as
$$t=\sqrt{\frac{2h}{g}}$$ $$t'=\sqrt{\frac{2h'}{g}}$$
where $h'=\frac{2}{3}h$. Then you can just construct the proportion $t'/t$, which is now the relative time spent in the first $2/3$... simply use physical intuition to subtract from 1 to get the relative time in the first $1/3$, and the "extra" answer isn't even a thing.