3
$\begingroup$

A rock is dropped into a well. A timer starts when the rock is dropped and is stopped when the noise of the rock hitting the ground of the well is heard. How deep is the well?

Here's what I have so far: $t = \left(\left(\frac{2\times h}{g}\right)^{\frac{1}{2}}+\left(\frac{h}{340}\right)\right)$ when I try to use the quadratic formula, I can express $h$, but when substitute different amounts for $t$, it doesn't seem right. I get a very small and a very large value for $h$. For example for $t=30\:\mathrm{s}$ I get $3\:\mathrm{m}$ and $42\:\mathrm{km}$, which is quite odd. I also tried it with the following set of equations:

$$\frac{1}{2}\left(g\times t_1^2\right)=h$$ $$t_2\times 340 \:\mathrm{m/s}=h$$ $$t_1+t_2=t$$

$\endgroup$
2

1 Answer 1

4
$\begingroup$

Let $v$ denote the speed of sound, and let's use your equations. Combine the second and third equations to get $$ \frac{h}{v} = t-t_1 $$ so that $$ t_1 = t-\frac{h}{v} $$ and therefore using the first equation $$ h = \frac{1}{2}g\left(t-\frac{h}{v}\right)^2 $$ Putting this in standard form to use the quadratic equation gives $$ h^2 - 2v\left(t+\frac{v}{g}\right)h + (vt)^2 = 0 $$ which gives $$ h = \frac{gtv+v^2\pm v\sqrt{v(2gt+v)}}{g} $$ for $g=9.8\,\mathrm{m}/\mathrm{s}^2$, $t=30\,\mathrm{s}$, adn $v=340\,\mathrm{m}/\mathrm s$, I get $2.5\,\mathrm {km}$ and $41.5\,\mathrm{km}$. The first solution is the correct one, for a 30 second fall, I think it seems quite reasonable; if you were freely falling for that amount of time without air resistance, you would travel about 4.4 kilometers.

$\endgroup$
4
  • $\begingroup$ Thanks, but the timer is stopped, when I hear the sound coming back from the well, how do you account for that? if you look carefully, at the end of my question, i also have this function. I need to somehow account for the time that the sound of the rock took to come back to me. $\endgroup$
    – Qqbt
    Feb 20, 2013 at 21:25
  • $\begingroup$ Oh sorry misread the question; solution edited. $\endgroup$ Feb 20, 2013 at 21:53
  • $\begingroup$ thanks, upvote, but can you unroll the square? because I did 7 (!) different ways of these starting equations and im always left with something different, could you take a look at Qmechanics solutions in the other question? how does that works out for him? $\endgroup$
    – Qqbt
    Feb 20, 2013 at 22:37
  • $\begingroup$ Ok I included another step. $\endgroup$ Feb 20, 2013 at 22:52

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .