College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 4 - Problems - Page 149: 37


The well is 46.1 meters deep.

Work Step by Step

Let $y$ be the depth of the well. Let $t$ be the time it takes the stone to fall. We can write an expression for $y$: $y = \frac{1}{2}at^2$ The time it takes for the sound to travel is $3.2-t$. We can write an expression for $y$: $y = 343~(3.2-t) = 1097.6-343t$ We can equate the two expressions for $y$ and solve for $t$: $\frac{1}{2}at^2 = 1097.6-343t$ $4.9t^2 +343t - 1097.6 = 0$ We can use the quadratic formula to find $t$: $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $t = \frac{-343 \pm\sqrt{(343)^2 - 4(4.9)(-1097.6)}}{(2)(4.9)}$ $t = -73.07~s, 3.066~s$ Since $t$ must be positive, $t = 3.066~s$. We can use $t$ to find $y$, the depth of the well: $y = \frac{1}{2}at^2$ $y = \frac{1}{2}(9.80~m/s^2)(3.066~s)^2$ $y = 46.1~m$ The well is 46.1 meters deep.
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