From the equation of motion for a freely falling body initially at rest, the distance $s$ traversed in time $t$ is given by $s = \frac{1}{2}gt^2$. Rearranging for $g$, we get $g = \frac{2s}{t^2}$. The maximum percentage error in $g$ is the sum of the percentage error in $s$ and twice the percentage error in $t$. Therefore, $\% \text{ error in } g = \% \text{ error in } s + 2 \times (\% \text{ error in } t) = e_1 + 2e_2$.