smc

Let $d$ be the depth of the well in feet, let $t_1$ be the number of seconds the rock took to fall to the bottom of the well, and let $t_2$ be the number of seconds the sound took to travel back up the well. We know $t_1+t_2=7.7$. Now we can solve $t_1$ for $d$: $$d=16t_1^2$$ $$\frac{d}{16}=t_1^2$$ $$t_1=\frac{\sqrt{d}}{4}$$ And similarly $t_2$: $$d=1120t_2$$ $$t_2=\frac{d}{1120}$$ So $\frac{d}{1120}+\frac{\sqrt{d}}{4}-7.7=0$. If we let $u=\sqrt{d}$, this becomes a quadratic. $$\frac{u^2}{1120}+\frac{u}{4}-7.7=0$$ $$u=\frac{-\frac{1}{4}\pm \sqrt{{\left(\frac{1}{4}\right)}^2-4(\frac{1}{1120})(-7.7)}}{\frac{2}{1120}}$$ $$u=560\cdot \left(-\frac{1}{4}\pm \sqrt{\frac{1}{16}+\frac{7.7}{280}}\right)$$ $$u=560\cdot \left(-\frac{1}{4}\pm \sqrt{\frac{36}{400}}\right)$$ $$u=560\cdot \left(-\frac{1}{4}\pm \frac{3}{10}\right)$$ We know $u$ is the positive square root of $d$, so we can replace the $\pm$ with a $+$. $$u=560\cdot \left(\frac{1}{20}\right)$$ $$u=\frac{560}{20}$$ $$u=28$$ Then $d=28^2=784$, and the answer is $\boxed{784\text{ ft.}}$.