Let's say an observer at the bottom of a well of depth $d$ which is situated on an incredibly large planet with gravitational force of $g$. We can assume that the value of $g$ remains constant regardless of depth. Let $c$ be the speed of sound on this planet.
Now consider an object that produces a sound at the instance it is dropped from the top of the well, falling toward the bottom of the well with acceleration of $g$. If the observer at the bottom sees the object hit the ground, and then $t$ seconds later they hear the sound produced by the object, how deep is the well in terms of $g,t$ and $c$?
Obviously this is a very artificial question because we are ignoring air resistance, yet sound waves need a medium in order to propogate. Even so, I think this is a cool little question that I've come up with, but I've become very stuck after struggling to solve this over the course of a few hours.
Note: I figure that the well must be very deep for the 'sonic boom' to occur, so we can assume that the planet has a radius that is more than the depth of the well.
I am only familiar with the $3$ main kinematics equations:
$$v=u+gt$$ $$v^2=u^2+2gd$$ $$d=ut+\frac{1}{2}gt^2$$
I'm fairly certain this is all you would need to solve this, but I can't seem to apply these equations to the question at hand because it is so abstract.
I would love to know the method of any solution to this question, because I'm very stuck.
Let's say that the object hits the bottom of the well after $t_0$ seconds. So, the depth of the well is given by $d = \frac 12 g t_0^2$. On the other hand, the depth of the well, measured in terms of the time that sound took to reach the bottom is given by $d = c(t+t_0)$. Hence, $$ \frac 12 gt_0^2 = c(t+t_0) \Leftrightarrow t_0 = \dfrac{c+\sqrt{c^2+2gct}}{g} $$
which implies that $$ d = \frac 12 g \dfrac{c+\sqrt{c^2+2gct}}{g} = \frac 12 \left(c+\sqrt{c^2+2gct}\right). $$