1951 AHSME Problems/Problem 39

Problem

A stone is dropped into a well and the report of the stone striking the bottom is heard $7.7$ seconds after it is dropped. Assume that the stone falls $16t^2$ feet in t seconds and that the velocity of sound is $1120$ feet per second. The depth of the well is:

$\textbf{(A)}\ 784\text{ ft.}\qquad\textbf{(B)}\ 342\text{ ft.}\qquad\textbf{(C)}\ 1568\text{ ft.}\qquad\textbf{(D)}\ 156.8\text{ ft.}\qquad\textbf{(E)}\ \text{none of these}$

Solution

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. $$\frac{u^2}{1120}+\frac{u}4-7.7=0$$ $$u=\frac{-\frac14\pm\sqrt{\left(\frac14\right)^2-4(\frac1{1120})(-7.7)}}{\frac2{1120}}$$ $$u=560\cdot\left(-\frac14\pm\sqrt{\frac1{16}+\frac{7.7}{280}}\right)$$ $$u=560\cdot\left(-\frac14\pm\sqrt{\frac{36}{400}}\right)$$ $$u=560\cdot\left(-\frac14\pm\frac3{10}\right)$$ We know $u$ is the positive square root of $d$, so we can replace the $\pm$ with a $+$. $$u=560\cdot\left(\frac1{20}\right)$$ $$u=\frac{560}{20}$$ $$u=28$$ Then $d=28^2=784$, and the answer is $\boxed{\textbf{(A)}}$.