Chapter 1.5, Problem 119E

Depth of a Well One method for determining the depth of a well is to drop a stone into it and then measure the time it takes until the splash is heard. If d is the depth of the well (in feet) and t₁, the time (in seconds) it takes for the stone to fall, then $d=16t_1^2$, so $t_1=\sqrt{d}/4$. Now if t₂ is the time it takes for the sound to travel back up, then d = 1090t₂ because the speed of sound is 1090 ft/s. So t₂ = d/1090. Thus the total time elapsed between dropping the stone and hearing the splash is

$t_1+t_2=\frac{\sqrt{d}}{4}+\frac{d}{1090}$

How deep is the well if this total time is 3 s?

To determine

The depth of the well if the total time is 3 s.

Answer to Problem 119E

The depth of the well if the total time is 3 s is $\underset{\_}{132.6\text{ft}}$.

Explanation of Solution

Given:

The total time elapsed between dropping the stone and hearing the splash is $t_1+t_2=\frac{\sqrt{d}}{4}+\frac{d}{1090}$ where, $t_1$ is the time takes for the stone to fall, $t_2$ is the time takes for the sound to travel back up and $d$ is the depth of the well.

Formula used:

Quadratic formula:

The solution of a quadratic equation of the form $a{x}^2+bx+c=0,a\ne 0$ can be obtained by using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

Total time elapsed between dropping the stone and hearing the splash is $3\text{s}$.

Substitute $3$ for $t_1+t_2$ in the above $t_1+t_2=\frac{\sqrt{d}}{4}+\frac{d}{1090}$ to obtain the depth of the well if the total time is 3 s.

$\begin{array}{l}3=\frac{\sqrt{d}}{4}+\frac{d}{1090}\\ 3=\frac{545\sqrt{d}+2d}{2180}\\ 2d+545\sqrt{d}=3\times 2180\\ 2d+545\sqrt{d}=6540\\ 2{\left(\sqrt{d}\right)}^2+545\sqrt{d}=6540\end{array}$

Let $x=\sqrt{d}$.Substitute $x$ for $\sqrt{d}$ in the above equation

$\begin{array}{l}2{\left(x\right)}^2+545x=6540\\ 2x^2+545x-6540=0\end{array}$

Use Quadratic formula to find value of x.

Substitute $2$ for $a$, $545$ for $b$ and $-6540$ for $c$ in the quadratic formula.

$\begin{array}{c}x=\frac{-545\pm \sqrt{{\left(545\right)}^2-4\times 2\times \left(-6540\right)}}{2\times 2}\\ =\frac{-545\pm \sqrt{297025+52320}}{4}\\ =\frac{-545\pm \sqrt{349345}}{4}\\ x=\frac{-545\pm 591.05}{4}\end{array}$

Simplify the above equation as follows.

$\begin{array}{l}x=\frac{-545+591.05}{4}\text{or }x=\frac{-545-591.05}{4}\\ x=\frac{46.05}{4}\text{or }x=-\frac{1136.05}{4}\\ x=11.512\text{or }x=-284.01\end{array}$

Note that, the distance can’t be negative.

Therefore, $x$ is equal to $11.512$.

Substitute $\sqrt{d}$ for $x$ in $x=11.512$ and solve for $x$.

$\begin{array}{l}\sqrt{d}=11.512\\ {\left(\sqrt{d}\right)}^2={\left(11.512\right)}^2\\ d=132.6\end{array}$

Thus, the depth of the well if the total time is 3 s is $\underset{\_}{132.6\text{ft}}$.