The pressure in a water spout and <PERSON>'s equation
This is a conceptual question about the application of Bernoulli's equation to a water spout.
There is a classical problem found in many physics texts which goes something like "you have a garden hose with a nozzle which flares inward so the radius is smaller at the end. How high does the water shoot into the air?"
So there are obviously details for the exact problem (like what is the angle of the hose, pressure or velocity, etc) but I am specifically interested in the applicability of Bernoulli's equation to the water which has left the hose. It would seem to be that after the water has left the hose, it is no longer satisfying the conditions for Bernoulli's equation. I can't quite put my finger on why; I can't exactly see the the flow is not laminar but since the pressure outside the flow (the air) is certainly not at the same pressure as the water, it's certainly doesn't seem to be steady.
In my understanding, you would treat the water in the nozzle with Bernoulli's equation (or continuity, depending on exactly conditions) and then simply treat the water as droplets acted on by gravity. If that's true, can someone clarify exactly what conditions of Bernoulli's equation are being violated?
Alternatively, if I am wrong, can you convince me that the water can still be considered a "fluid" for the purposes of applying Bernoulli's equation?
EDIT: A specific example in response to a comment. 1018
This shows that assuming <PERSON> still applies is equivalent to assuming the pressure of the stream is the same as the atmospheric pressure. Vertical oil pipe of height $h_1$, oil (gauge) pressure at its base of $P$ and fluid velocity $v$. How high into the air down the oil shoot?
Solution 1) <PERSON> fails when the fluid exits the pipe. The pressure when the fluid leaves is equal to 0 (gauge) so we have <PERSON>'s equation at the top of the pipe
$$P+\frac{1}{2}\rho v^2=\frac{1}{2}\rho v'^2+\rho g h_1\longrightarrow v'^2=v^2+\frac{2P}{\rho}-2gh_1$$ Then using energy conservation we have $$mgh_1+\frac{1}{2}mv'^2=mgh_2\longrightarrow h_2=h_1+\frac{v'^2}{2g}$$ $$h_2=h_1+\frac{1}{2g}\left( v^2+\frac{2P}{\rho}-2gh_1 \right)=\frac{v^2}{2g}+\frac{P}{g\rho}$$.
Solution 2) <PERSON> holds throughout the motion. The (gauge) pressure at the very top (and throughout the entire spout) is zero, so <PERSON> gives us $$P+\frac{1}{2}\rho v^2=\rho g h_2\longrightarrow h_2=\frac{P}{g\rho}+\frac{v^2}{2g}$$ 707