# 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 $\boxed{\textbf{(A)}}$.

## See Also

 1951 AHSC (Problems • Answer Key • Resources) Preceded byProblem 38 Followed byProblem 40 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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