Problem 13.40

13.40.
Let d be the distance that the stone drops. The time it takes for the stone to hit the bottom is given by
$d = \frac{1}{2} g t^{2}$
$t = \sqrt{\frac{2d}{g}}$
The time that it takes for the sound to travel back to the top is given by
$t = \frac{d}{v_{s}}$
So, the total time that it takes to hear the stone hit is given by
$t = \frac{d}{v_{s}} + \sqrt{\frac{2d}{g}}$
We can multiply through to get a quadratic in $\sqrt{d}$.
$d + \sqrt{\frac{2}{g}} v_{s} \sqrt{d} - v_{s} t ~=~ 0$
From the quadratic equation, we find that
$\sqrt{d} = -\frac{1}{2} (\sqrt{\frac{2}{g}} v_{s} \pm \sqrt{\frac{2v_{s}^{2}}{g} + 4v_{s}t})$
Plugging in the numbers, we find that
$\sqrt{d} = \frac{1}{2}(-155.0 \pm \sqrt{38,000})$
Choosing the positive root, we have that
$\sqrt{d} = 20.0$
d = 400 m
Now, if we ignore the speed of sound, $t = \sqrt{\frac{2d'}{g}}$
$d' = \frac{1}{2} gt^{2}$
d' = 510 m
The percentage error is given by
$error = \frac{d' - d}{d}$
error = .275
Percentage error $= 27.5 \%$