Stone Falls Into well, anyone know solution?

Question

Stone falls into a deep well, you hear it hit 3.2 seconds later. Thats the time it took for the stone to fall + the time it took for sound to travel back up the well to you. Sound is traveling at 343m/s how deep is the well?


There werent any other givens and I cant think of a way to approach the problem without knowing more, any ideas?

Answer 1

Gravitational acceleration g is 9.8 m/s^2, so let d = depth of well. The time of the fall is figured as follows:

d = 1/2 g t^2
t = √ (2 d / g)

So that the total elapsed time is:

3.2 = √ (2 d / 9.8) + d / 343 (everything in meters and seconds). We solve for d:

( 3.2 - d/343 )^2 = (2 d / 9.8), leading to a quadratic equation, with the solution:

d = 46.1 meters To check, we find the time of fall:

t1 = √ (2 * 46.1 / 9.8) = 3.07, and time of sound:

t2 = 46.1 / 343 = 0.13, so that

t1 + t2 = 3.2

Answer 2

NICE PROBLEM. The well is just over 46 m 5 cm deep!

{{But see my "SALUTORY ADMONITION" on the limited number of justifiable figures to which this answer can really be given, among my postscripts below, at ***.}}

Here is how to do this problem.

Let the stone fall for ' t ' seconds; in that time, it falls a distance (depth) d :

d = (1/2) g t^2 metres.

The sound travels back for (3.2 - t) seconds, travelling a distance:

d = 343 (3.2 - t) metres.

These two distances necessarily being equal, one has:

(1/2) g t^2 = 343 (3.2 - t) [provided units are consistent, which they are, with g = 9.8 m/sec^2 !] Therefore:

(1/2) (9.8) t^2 = 343 (3.2 - t), or:

4.9 t^2 +343 t - 343 (3.2) = 0, or, dividing by 4.9, (!) :

t^2 + 70 - 224 = 0. The solutions to this are:

t = [ - 70 +/- sqrt (70^2 + 4x224)] / 2 = [ - 70 +/- sqrt (5796)] / 2, i.e.

[ - 70 +/- 76.131465... ] / 2 = 3.065733... secs. (The negative one is unphysical.)

Now check the depth, both ways:

(1/2) g t^2 = 46.0537... m ; 343 x (3.2 - t) = 46.0537... m. THEY CHECK!

So : The well is just over 46 m 5 cm deep!

Live long and prosper.

P.S. I have a feeling I could have worked this with ' d ' as the variable to obtain directly, and may return to do that later --- family problems call, now.

Later Postscript: ' theyuks ' forgot that the distance travelled under constant acceleration is (1/2) g t^2 --- his expression lacked the (1/2) --- so his (initial) answer is wrong. What's more, he didn't give you the actual depth of the well, which was what you really wanted.

*** SALUTARY ADMONITION: The number of justifiable significant figures.

In my answer I dropped my customary caution and said that the depth of the well (D here) would be ~ 46m 05cm, while scythian1950 (a public slave or watchman?) determined it to be 46.1m. NEITHER figure is in fact justified.

Because the total elapsed time from dropping the stone to hearing it plop into the water is given to a mere 2 significant figures, the only JUSTIFIABLE conclusion (at best!) is that the well is approximately 46m deep. In other words, with limited two significant figure input for the total time ' t ' (not to mention the value g = 9.8 m/sec^2 that both of us also assumed), the answer can't really be given to more than two significant figures.

Indeed, if one examines this a bit more closely, the situation is slightly worse. This particular "falling stone/returned sound" problem is not really well-conditioned, in the sense that the time spent falling is ~ 96 % of ' t ', while the sound travel time is only ~ 4 % of t. That means that (delta D) / D is almost (or " ~ ") 2 (delta t) / t, so that the uncertainty in ' D ' due to rounding in ' t ' is almost TWICE what one would naively expect.

This is confirmed by calculating the range of ' D ' that could follow from the allowed rounding. With only two significant figures given, ' t ' could lie in the range 3.150000... to 3.250000... seconds if unlimited accuracy were allowed and available for ALL the given inputs (i.e. just extend the given inputs by 0000... etc. indefinitely). Such unlimited accuracy would yield a range of possible depths from 44.68202... - 47.44445... metres.

As can be seen from these results, the total range of uncertainty in the depth arising from the rounding of the total time ' t ' is ~ 0.05998... of the centrally calculated value. This is almost TWICE the ~ 1/32 or ~0.03125 naively expected from the essentially LINEAR expectations upon which the statement "number of significant output figures = number of significant input figures" is based.

This situation also manifests itself more obviously (though scythian1950 din't mention this) in what one might consider the "more direct" way of calculating ' D ' from its own quadratic equation, rather than from first evaluating t_fall and then t_sound. The solution for ' D ' from "D^2 - ... = 0," for inputs of unlimited accuracy, involves [26205.2... +/- 26113.1...] / 2 or 92.1 / 2. This means that the solution is the DIFFERENCE between two numbers that differ from one another by only ~ 0.0035 of either of them. Now THAT's a CLEAR indication of ill-conditioning !

The bottom line is: we can anly assert that the depth of the well is appproximately 46 metres, and the uncertainty in this estimate is about +/- 1.4 metres.

Answer 3

Well, you can guesstimate based on what you know of physics:
Gravitational acceleration = 9.8 m/s/s
Speed of sound (S) at sea level= 344 m/s

With a relatively short fall and (we assume) a reasonably small stone, air resistance is effectively zero

So T(total) = T(fall) + T(sound), and if you work it out, you get
Height(t) = -9.8 x t(fall)^2 +v(0)t(fall) + height(0)
And h = S x T(sound)

Assume v(0) = 0 m/s and that term goes away. So now height(t) = -9.8 x t(fall)^2 + S x t(sound). But since you know that t(total) = 3.2 sec, you can rearrange to get t(fall) = 3.2 sec - t(sound)

Substitute, and you get h(t) =-9.8 X t(fall)^2 - S x t(fall) + 3.2S

So now, you have the equation in terms of a single variable. Now, at the bottom of the well, h(t) = 0, right? It's the bottom? So now you've got a quadratic equation, and can solve for t(fall)

t(fall) = - S +/- SQRT(S^2 - 4(-9.8)(S x 3.2)) / 2 (-9.8)
I'll simplify and you get -344 +/- 401.855 / 19.6
You can't have negative time (well, not for practical purposes, anyway), so you have to add instead of subtract, and you get t(fall) = 2.95 seconds, which then makes your t(sound) = 0.25 seconds. If you plug the numbers back into your original equations, it works out pretty well (I rounded in a couple of places, so it won't come out perfectly, but the process is valid.)

Answer 4

Not sure which formula you want to use. There are many ways to calculate it.
But we do know that with gravity, an object falls at approx 4.9m/s