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 (ProblemsAnswer KeyResources)
Preceded by
Problem 38
Followed by
Problem 40
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