Is the loudness of a ball that you drop a measure of the amount of energy you put in to lift it? W

Bottom Line Up Front
This is a complicated problem, but with some simplifying assumptions and some (hopefully) reasonable guesses, I think the sound amplitude will increase about 6 dB for a doubling of drop height, and the sound energy is (very) roughly proportional to the square of the kinetic energy of the ball dropping.
Discussion
Sound is a mechanical wave through a medium, which in this case is air. There are two common sources of sound energy: a directional forcing of the air, or an omni-directional expansion/compression of the air. While I am have not studied this particular sound problem in detail, I would expect the dominant sound source to be the compressive type as the ball compresses the air between it and the ground. Other sound sources include the elastic response of the ball (maybe dominated by the directional forcing?) and the sound of the ball falling (comes from the turbulent wake and is actually a different source type than the two I described above). Let us consider a perfectly inelastic ball (to ignore the elastic response) that is shaped like a pancake (to get rid of some of the hairy geometry that complicates our understanding).
As the ball (OK, it is really a pancake) falls and hits the ground, the air underneath it needs to get out of the way. If the impact happens slowly enough [e.g., you gently slide the pancake onto the surface (I am imagining making a classic American breakfast with a spatula, at this point)], the air motion can be laminar and very little compression will happen. That compression that does occur and generates sound waves will have a very low frequency which, for complicated reasons I won't go into here, won't propagate and won't be audible. As the pancake falls faster the air needs to move faster and will start to compress more, leading to audible sound generation. I am purely guessing here (again, I have not studied this problem in depth), but my intuition from fluid mechanics suggests that the linear rate of compression is proportional to the square of the falling velocity after some critical sound-generating velocity, or
$p_{\text{out}}=A(v-<!--\; -\; -->v_0{)}^2\equiv <!--\; \equiv \; -->A\mathrm{\Delta <!--\; \Delta \; -->}{v}^2,$
where $p_{\text{out}}$ is the pressure generated at the edge of the pancake, $A$ is some constant, $v$ is the pancake fall velocity, $v_0$ is the critical pancake velocity, and $\mathrm{\Delta <!--\; \Delta \; -->}v=v-<!--\; -\; -->v_0$ is the velocity above the critical velocity. This expression would only be valid for $v\ge <!--\; \ge \; -->v_0$
Assuming that my guess in the previous paragraph is correct, we can now talk about the loudness. Human perception of sound is actually very complicated , but we will use the decibel level as a useful and simple surrogate. The level of a sound wave in air is given by
$L=20{\log }_{10}<!--\; \; -->(p/p_{\text{ref}}),$
where $p$ is the pressure amplitude and $p_{\text{ref}}$ is a reference pressure of 20~$\mathrm{\mu <!--\; \mu \; -->}$Pa. Substituting the above expression then yields
$L=20{\log }_{10}<!--\; \; -->\left(A/p_{\text{ref}}\right)+40{\log }_{10}<!--\; \; -->(\mathrm{\Delta <!--\; \Delta \; -->}v).$
Ignoring the effects of air resistance on the fall velocity, we can estimate $v$ from the interchange of kinetic and potential energy, such that $v\approx <!--\; \approx \; -->\sqrt{2gh}$, where $g$ is the gravitational acceleration and $h$ is the height of the drop. For drops sufficiently high that $v\gg <!--\; \gg \; -->v_0$, we can then approximate
$L\approx <!--\; \approx \; -->C+20{\log }_{10}<!--\; \; -->(2gh),$
where $C$ is some constant. Thus, doubling the height would increase the level by $20{\log }_{10}<!--\; \; -->(2)\approx <!--\; \approx \; -->6$ dB.
Finally, some comments on sound energy. For a purely propagating plane wave (and for spherical waves sufficiently far from the origin) the acoustic energy is proportional to the square of the pressure. In our case here we find that
$p^2=A^2\mathrm{\Delta <!--\; \Delta \; -->}{v}^4\approx <!--\; \approx \; -->A^2{v}^4\propto <!--\; \propto \; -->E_k^2,$
where $E_k$ is the kinetic energy of the pancake. Thus, the sound energy is proportional to the square of the kinetic energy.

The loudness is more tied to the loss of energy. When the ball hits the ground, especially if it's a relatively inelastic type of ball or badly pumped, or if the floor is inelastic, mechanical energy is released, some of it as sound