Water rocket physics According to this page, the current record for greatest altitude achieved by a water and air propelled rocket is 2044 feet (623 meters).
In this connection, i had one question: How the  greatest altitude depends of the density of the liquid used in the rocket?
 Suppose that we have the opportunity to choose a liquid with an arbitrary density and with the same hydrodynamic properties as water. Then there must be a density at which the greatest altitude has its absolute maximum, i think.   
To be more specific let's consider a simpler problem:  
Let a two-liter soda bottle rocket, filled with some amount of a liquid mentioned above and air at the atmospheric pressure, is launched in deep space($g=0$)(say an astronaut is experimenting). What is the absolute maximum of the speed the rocket can achieve? For simplicity let's assume the expansion of the air to be isothermal (which it is not).

 A: This problem is a combination of fluid mechanics and the physics of the rocket equation from Newton’s laws.  The two need to be put together properly.
To derive the rocket equation suppose there exists a mass $M$ that fragments into $M~-~\delta M$ and $\delta M$, where $\delta M~<<~M$. Further, suppose that $\delta M$ flies away at a velocity $V$.  The large mass will then experience a change in velocity $-\delta v$ so that
$$
0~=~\delta MV~-~(M~-~\delta M)\delta v.
$$
If these increments are small the term $\delta M\delta v$ may be ignored.  This is the conservation of momentum for a brief increment in the rocket flight. From here in the calculus limit these increments become infinitesimal
$$
0~=~VdM~-~Mdv,~\rightarrow~\int_0^v dv^\prime~=~V\int_M^m {{dM^\prime}\over{M^\prime}}.
$$
This leads to a final velocity of the rocket with mass $m$ after expending a mass $M~-~m$ of burned fuel plus oxidant $v~=~Vln(M/m)$.  This equation is not adequate for the velocity of the reaction mass is assumed to be constant.  
We consider the physics internal to the rocket bottle.  Air pressure pushes the water out, where this pressure depends on the volume of the air.  In fact this is the natural gas law $pV~=~NkT$, and $p~=~NkT/V$.  We also need the Navier Stokes equation in one dimension.  We then have
$$
\rho\frac{dv}{dt}~=~\frac{dp}{dx}
$$
The pressure is dependent on the volume with $V~=~2\pi R^2 x$ and we have that
$$
\frac{dv}{dt}~=~-\frac{NkT}{2\pi R^2\rho x^2}
$$
The volume of water decreases as $x$ increases, as the air expands.  This $x~=~ut$ will be related to the velocity of the water jet out the nozzle by the ratio of the bottle area and nozzle area $2\pi r^2$ so $u~=~v(r/R)^2$.  Thus 
$$
\frac{dv}{dt}~=~-\frac{NkTR^2}{2\pi v^2\rho r^4t^2}
$$
and thus by rearranging things and solving the integrations
$$
\frac{v^3}{3}~=~\frac{NkTR^2}{6\pi r^4\rho t}
$$
and so the water velocity out the nozzle is
$$
v~=~t^{-1/3}\Big(\frac{NkTR^2}{2\pi r^4\rho}\Big)^{1/3}
$$
I have been a bit cavalier here, for one needs to take care of the end points of this integration to cut off the integral at some time t.  Clearly the water velocity does not asymptote to zero as time goes to infinity.  I leave that detail to the reader.
Now the job is to rework the rocket equation with this as the plume velocity $V$.  One also expresses the $dM$, the change in the mass, according to the rate of change in the column of water in the bottle.  At this point this should be reduced to further calculations of this sort.  I will halt this at this point, for this is already rather TeX intensive and pages tend to be slow.
