A climber wants to know the depth of a well. He drops a rock and 4 seconds later he hears a noise. The speed of sound is 340 m/s.

Okay, so

$ t= 4s $

But t has 2 parts,

$ t = t_1 + t_2 $

Because there is a time when the object is hitting the bottom of the well, and then another time which is sound going back to his ear.

I used formula:

$\Delta x = V_{ox} t + (1/2)a_x t^2 $

to get height in terms of $t_1$,

So he dropped the object from the top, its initial velocity = 0

thus I'm left with

$H_{t1} = (1/2)(9.8m/s^2)(t_1^2)$

$H_{t1} = (4.9m/s^2)(t_1^2)$

Then I set up another equation for height. The reason I'm doing this is to set my two height equations = to each other and isolate my $t_1$ variable because I need it to complete my problem

The reason I used 1360m - 340 m/s * $t_1$ is because 4 seconds times 340 m/s is = to 1360, and im subtracting one of the times multiplied by the speed of sound to get the height of the well.

$H_{t2} = 1360 m - (340m/s \times t_1) $

Okay so here is my first question:

Is everything that I'm doing right now making sense? I'm not asking if this is the most efficient way to do it, I'm asking is my method right now sound, because this is what I would do on a test...

Anyways, I set the equations =

$t_1 = 4s - t_2 $

$(4.9m/s^2)(4s - t_2)^2 = 1360 m - (340m/s \times t_1) $

$(4.9m/s^2) \times (16s^2 -8s (t_2) + (t_2)^2) $ = $1360 m - (340m/s \times t_1) $

$78m - 39\frac ms(t_2) + 4.9 \frac m{s^2}(t_2)^2 = 1360 m - (340m/s \times t_1) $

$1360 m - (340m/s \times t_1) = 78m - 39\frac ms(t_2) + 4.9 \frac m{s^2}(t_2)^2 $

$1360m - (340 \frac ms \times (4s-t_2)) = 78m - 39\frac ms(t_2) + 4.9 \frac m{s^2}(t_2)^2 $

$1360m - (1360m - 340\frac ms t_2) = 78m - 39\frac ms(t_2) + 4.9 \frac m{s^2}(t_2)^2 $

$-340\frac ms t_2 = 78m - 39\frac ms(t_2) + 4.9 \frac m{s^2}(t_2)^2 $

$ 0 = 78m + 301\frac ms(t_2) + 4.9 \frac m{s^2}(t_2)^2 $

using quadratic equation I solve and get $t_2 = -.26$, but I use the absolute value here and get $t_2 = .26 $

Now I find that $t_1 = 3.74 seconds $

and use $V_x = a_x t $ because initial velocity was zero and get$ V_x = 36.7$

now finally to get height I use $\Delta x = (1/2)(V_x)t $ again $V_ox$ = 0 so I don't count it and get

$height of well = 68.6 meters$

What did I do wrong though? Does my method seem sound for everyone reading? I think I did this problem perfectly fine but according to:


the answer is 70.48 m.

Did maybe I just have a calculation error but my method was correct? On a test do you think a teacher would still give me credit if I was only off by 1.88 meters? Can anyone please help because this problem is driving me crazy.

  • $\begingroup$ You had two equations for $H_t1$ and $H_t2$ both in terms of $t_1$. If you set them equal to each other you then have a quadratic equation in $t_1$. Solve that then use it to find $H$. ... Instead of doing that you made 2 complicated and totally unnecessary substitutions, replacing $t1$ by $t2$. It is not surprising you made a mistake somewhere. $\endgroup$ Commented Feb 14, 2017 at 3:06

2 Answers 2


By the time you're putting this much effort into a problem, you should solve it algebraically, then plug in your values at the end. That allows you to sanity-check your answer by finding the limiting behavior and dimensional analysis.

For example, what if the speed of light goes to infinity? Then the calculated height should just be $c \cdot t$, where $c$ is the speed of sound, and $t$ is the time delay for the noise. Does your equation give you that for $c\rightarrow \infty$? What if $g \rightarrow \infty$? Do all the terms in your expression for $h$ have units of length? If not, go back and find the mistake.

Solving problems algebraically can also lead to lots of interesting insights about the problem that you won't get if you just mess around with numbers the whole time. It also prevents inaccuracies due to compounding rounding errors.

Here's how I'd start:

$$h = \frac 1 2 g t_A^2$$ $$h = c t_B$$ $$t = t_A + t_B$$

Solve for $h$, and don't plug in a single number until you have a simplified algebraic expression for $h$

  • 1
    $\begingroup$ Thank you. I will report back. I did not think to do it this way, though this seems very efficient, and doesn't require me to change my method. $\endgroup$
    – Nak Leng
    Commented Feb 13, 2017 at 2:48
  • $\begingroup$ Great - I'd be glad to help if you run into trouble with the algebra. $\endgroup$
    – Brionius
    Commented Feb 13, 2017 at 3:06
  • $\begingroup$ Hey I am having trouble with the algebra sadly.... is there anyway I can message you to discuss? I can't seem to find a message button on your profile. @Brionius $\endgroup$
    – Nak Leng
    Commented Feb 14, 2017 at 19:21
  • $\begingroup$ @Nak I'd be happy to help. The easiest thing would be for you to create a new questions stating that you're trying to do this algebraically, post what you've done so far and what's confusing, and we can help you get past whatever the sticking point is. $\endgroup$
    – Brionius
    Commented Feb 14, 2017 at 20:16
  • $\begingroup$ Thanks. I'm going to work on it for another hour or so and if nothing pops up and I'm still stuck I will write another comment with a link to the question on here. Thank you so much, like you have said there are a lot of insights to this problem by doing it algebraically! $\endgroup$
    – Nak Leng
    Commented Feb 14, 2017 at 20:21

You were off to a good start, but you got bogged down. The equation $H=\frac{1}{2}9.8 t_1^2$ was correct. That left $4-t_1$ seconds for the sound to return back over the distance H. So $H=340(4-t_1)$. The two H's are the same, so just eliminate them and solve for $t_1$.


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