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Difference between revisions of "1951 AHSME Problems/Problem 39"

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==Problem==
 
==Problem==
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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:
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<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>
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==Solution==
 
==Solution==
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Unsolved
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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>:
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<cmath>d=16t_1^2</cmath>
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<cmath>\frac{d}{16}=t_1^2</cmath>
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<cmath>t_1=\frac{\sqrt{d}}4</cmath>
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And similarly <math>t_2</math>:
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<cmath>d=1120t_2</cmath>
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<cmath>t_2=\frac{d}{1120}</cmath>
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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.
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<cmath>\frac{u^2}{1120}+\frac{u}4-7.7=0</cmath>
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<cmath>u=\frac{-\frac14\pm\sqrt{\left(\frac14\right)^2-4(\frac1{1120})(-7.7)}}{\frac2{1120}}</cmath>
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<cmath>u=560\cdot\left(-\frac14\pm\sqrt{\frac1{16}+\frac{7.7}{280}}\right)</cmath>
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<cmath>u=560\cdot\left(-\frac14\pm\sqrt{\frac{36}{400}}\right)</cmath>
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<cmath>u=560\cdot\left(-\frac14\pm\frac3{10}\right)</cmath>
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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>.
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<cmath>u=560\cdot\left(\frac1{20}\right)</cmath>
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<cmath>u=\frac{560}{20}</cmath>
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<cmath>u=28</cmath>
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Then <math>d=28^2=784</math>, and the answer is <math>\boxed{\textbf{(A)}}</math>.
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== 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 $7.7$ 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 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

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