Height of water fountain due to pressure difference This was a question on a fluid dynamics exam: the pressure $p_A$ on a plane is lower than the atmospheric pressure. We fill a water bottle with a straw on the ground and open it in the plane. Water will come out due to the pressure difference. Calculate the maximum height of the water fountain above the straw.
What I think happens is that the only place where water can come out is at the straw, so we have a force upwards due to the pressure difference $p_{bottle}-p_A$. Water will only move upwards if this force can counteract the gravitational force $\rho_{water} g$ (I think we can negate the fact that we are high up and the gravitational force is weaker). Then I tried to calculate it as if it was a point mass thrown up with a force $p_{water}-p_A$. But to be honest I'm not even able to calculate the initial velocity (I'm notoriously bad at physics). 
Any help or insight will be appreciated.
 A: Use Bernoulli's Principle: the energy equation for incompressible, inviscid fluid flow:
$$p_{bottle}+\frac12 \rho v_{bottle}^2+\rho gh_0=p_A+\frac12 \rho v_1^2+\rho gh_1$$
Because the bottle is much wider than the straw and with the continuum equation:
$$A_{bottle}v_{bottle}=A_{straw}v_{straw}$$
Where the $A$ are cross-sections and with:
$$\frac{A_{straw}}{A_{bottle}}<< 1$$
So that $v_{bottle}\approx 0$.
Now as the water flows out of the straw at $v_{straw}$ it will decelerate due to gravity, until $v_1=0$ and it has reached the height $h_1$ (the water then starts falling back to Earth).
Plug it all in and we have:
$$p_{bottle}+\rho gh_0\approx p_A+\rho gh_1$$
Or:
$$\Delta h\approx \frac{p_{bottle}-p_A}{\rho g}$$
In essence what happens here is that pressure energy is converted to potential energy.
A: In my judgment, it is invalid to apply the Bernoulli equation to this problem because the Bernoulli equation applies only to steady state flows (or to flows that are nearly steady state), and this problem involves neither.  However, water is very nearly incompressible, particularly for a change in pressure that is relatively small (like in the present case).  So the volume of water will not change significantly when the pressure on it is reduced.  That means that, for all practical purposes, there will be no fountain effect.  All that will happen will be that the pressure in the water within the bottle will drop to the new pressure, essentially instantaneously.
