Difference between revisions of "1951 AHSME Problems/Problem 39"
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| + | ==Problem== | ||
| + | |||
| + | A stone is dropped into a well and the report of the stone striking the bottom is heard <math>7.7</math> seconds after it is dropped. Assume that the stone falls <math>16t^2</math> feet in t seconds and that the velocity of sound is <math>1120</math> feet per second. The depth of the well is: | ||
| + | |||
| + | <math> \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} </math> | ||
| + | |||
| + | ==Solution== | ||
| + | Let <math>d</math> be the depth of the well in feet, let <math>t_1</math> be the number of seconds the rock took to fall to the bottom of the well, and let <math>t_2</math> be the number of seconds the sound took to travel back up the well. We know <math>t_1+t_2=7.7</math>. Now we can solve <math>t_1</math> for <math>d</math>: | ||
| + | <cmath>d=16t_1^2</cmath> | ||
| + | <cmath>\frac{d}{16}=t_1^2</cmath> | ||
| + | <cmath>t_1=\frac{\sqrt{d}}4</cmath> | ||
| + | And similarly <math>t_2</math>: | ||
| + | <cmath>d=1120t_2</cmath> | ||
| + | <cmath>t_2=\frac{d}{1120}</cmath> | ||
| + | So <math>\frac{d}{1120}+\frac{\sqrt{d}}4-7.7=0</math>. If we let <math>u=\sqrt{d}</math>, this becomes a quadratic. | ||
| + | <cmath>\frac{u^2}{1120}+\frac{u}4-7.7=0</cmath> | ||
| + | <cmath>u=\frac{-\frac14\pm\sqrt{\left(\frac14\right)^2-4(\frac1{1120})(-7.7)}}{\frac2{1120}}</cmath> | ||
| + | <cmath>u=560\cdot\left(-\frac14\pm\sqrt{\frac1{16}+\frac{7.7}{280}}\right)</cmath> | ||
| + | <cmath>u=560\cdot\left(-\frac14\pm\sqrt{\frac{36}{400}}\right)</cmath> | ||
| + | <cmath>u=560\cdot\left(-\frac14\pm\frac3{10}\right)</cmath> | ||
| + | We know <math>u</math> is the positive square root of <math>d</math>, so we can replace the <math>\pm</math> with a <math>+</math>. | ||
| + | <cmath>u=560\cdot\left(\frac1{20}\right)</cmath> | ||
| + | <cmath>u=\frac{560}{20}</cmath> | ||
| + | <cmath>u=28</cmath> | ||
| + | Then <math>d=28^2=784</math>, and the answer is <math>\boxed{\textbf{(A)}}</math>. | ||
| + | |||
== See Also == | == See Also == | ||
{{AHSME 50p box|year=1951|num-b=38|num-a=40}} | {{AHSME 50p box|year=1951|num-b=38|num-a=40}} | ||
Latest revision as of 12:59, 19 April 2014
Problem
A stone is dropped into a well and the report of the stone striking the bottom is heard 
seconds after it is dropped. Assume that the stone falls 
feet in t seconds and that the velocity of sound is 
feet per second. The depth of the well is:
Solution
Let 
be the depth of the well in feet, let 
be the number of seconds the rock took to fall to the bottom of the well, and let 
be the number of seconds the sound took to travel back up the well. We know 
. Now we can solve for :
![\[d=16t_1^2\]](http://latex.artofproblemsolving.com/c/d/2/cd2a5fec1f25b11a7ea99e02fce3a42d8ba4de73.png)
![\[\frac{d}{16}=t_1^2\]](http://latex.artofproblemsolving.com/b/c/4/bc46148684a7a6a6e76fc59bef32a71975bc4ff5.png)
![\[t_1=\frac{\sqrt{d}}4\]](http://latex.artofproblemsolving.com/d/9/6/d962f9e81da97da1b218196bd4d0b7a31188fcd0.png)
And similarly :
![\[d=1120t_2\]](http://latex.artofproblemsolving.com/1/e/3/1e35f7f7c4254298729bb6607a17c17d0b4eae50.png)
![\[t_2=\frac{d}{1120}\]](http://latex.artofproblemsolving.com/8/4/8/848a7dd43ac97f5d2f3c9146d6130173740e933d.png)
So 
. If we let 
, this becomes a quadratic.
![\[\frac{u^2}{1120}+\frac{u}4-7.7=0\]](http://latex.artofproblemsolving.com/3/4/2/3427ff46b91ac2735ebd93fe1ff59db88b4749ce.png)
![\[u=\frac{-\frac14\pm\sqrt{\left(\frac14\right)^2-4(\frac1{1120})(-7.7)}}{\frac2{1120}}\]](http://latex.artofproblemsolving.com/f/6/c/f6ce6927c41b22bc1c0a357fb8374274adc1e96f.png)
![\[u=560\cdot\left(-\frac14\pm\sqrt{\frac1{16}+\frac{7.7}{280}}\right)\]](http://latex.artofproblemsolving.com/9/8/4/984c113b985348cea1e8e53565d257ce3fc153a0.png)
![\[u=560\cdot\left(-\frac14\pm\sqrt{\frac{36}{400}}\right)\]](http://latex.artofproblemsolving.com/9/d/f/9df4945ef343c0a12b48d61ac95cc1f244dc2b0a.png)
![\[u=560\cdot\left(-\frac14\pm\frac3{10}\right)\]](http://latex.artofproblemsolving.com/0/a/4/0a4d0b0aa1cccb75fd57a6bfd601262b07d63014.png)
We know 
is the positive square root of , so we can replace the 
with a 
.
![\[u=560\cdot\left(\frac1{20}\right)\]](http://latex.artofproblemsolving.com/2/6/8/2687c6a4efddeec600052d4bda65184426086b96.png)
![\[u=\frac{560}{20}\]](http://latex.artofproblemsolving.com/6/8/f/68f16dad7885d9f3f1b2b5c8ff9526b9e06ef0dd.png)
![\[u=28\]](http://latex.artofproblemsolving.com/4/4/6/4466d6e82a14095b8b87e275ad36d824ffd20fc0.png)
Then 
, and the answer is 
.
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 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. 
