Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 12 - Sound - Problems - Page 354: 7

Answer

$h = 33~m$

Work Step by Step

Let $h$ be the height of the cliff. Let $T$ be the total time it takes until we hear the sound. Let $t_1$ be the time that the stone drops. $h = \frac{1}{2}gt_1^2$ Let $t_2$ be the time that the sound of the splash travels up the cliff. Note that $t_1+t_2 = T$ $h = v~t_2$ $h = v~(T-t_1)$ $h = v~T-v~t_1$ We can equate the two expression for $h$ and solve for $t_1$. $\frac{1}{2}gt_1^2 = v~T-v~t_1$ $\frac{1}{2}gt_1^2 +v~t_1-v~T = 0$ $\frac{1}{2}(9.80~m/s^2)t_1^2 +(343~m/s)~t_1-(343~m/s)(2.7~s) = 0$ $(4.90~m/s^2)t_1^2 +(343~m/s)~t_1-(926.1~m) = 0$ We can use the quadratic formula to find $t_1$. $t_1 = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$ $t_1 = \frac{-(343)\pm \sqrt{(343)^2-(4)(4.9)(-926.1)}}{(2)(4.9)}$ $t_1 = 2.6~s, -73~s$ Since the negative value is unphysical, the solution is $t_1 = 2.6~s$. We can find the height of the cliff. $h = \frac{1}{2}gt_1^2$ $h = \frac{1}{2}(9.8~m/s^2)(2.6~s)^2$ $h = 33~m$
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