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 $h$
By the Sakur-Tetrode equation, the entropy of a monatomic, ideal gas is given by
$$\frac{S}{k_BN}=\ln\left[\frac VN\left(\frac{4\pi m}{3h^2}\frac{U}{N}\right)^{3/2}\right]+\frac52$$
For our purposes, it will make sense to use the ideal gas law to express $V$ in terms of $P$ and $T$, and to express $U=\frac32Nk_BT$, so we get
$$\frac{S}{k_BN}=\ln\left[\frac {k_BT}{P}\left(\frac{2\pi m}{h^2}\cdot k_BT\right)^{3/2}\right]+\frac52$$
So, as we can see, for a constant pressure $P$, the entropy of the ideal gas is a monotonically decreasing function with respect to decreasing $T$; if we decrease $T$, we decrease $S$.
I suspect your confusion comes from thinking that $S$ can never decrease, but this is only the case for isolated systems. If you are forcing an ideal gas to undergo an isobaric compression, then the system is no longer isolated, and so the entropy can decrease (entropy will increase elsewhere though).
As a separate argument, the entropy is a state function, meaning its value only depends on the state, not how you got there. Now, let's consider your isobaric process, and let's say we do an isobaric expansion and then an isobaric compression back to the original state (this is the original state of the system, not the original state of the system as well as the surroundings, which is impossible to achieve). Since entropy is a state function, the entropy ends at where it started. But this means one of two things happened
- The entropy remained constant the entire time
- The entropy increased as well as decreased during this process.
If you showed that the change in entropy is non-zero for some part of this, then you have to conclude that it is possible to decrease the entropy of an ideal gas. Also, note that this argument is not dependent on use of an ideal gas specifically.
Best Answer
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.