Distance traveled by a water jet I helped my kid in a science fair project, where we punctured holes in a water bottle at various heights and then measured the distance traveled by the water jets before they hit the ground. The experimental observation is that the plot of distance traveled to height of hole "appears" parabolic with maximum distance traveled by the almost the "center" jet.  I want to have a theoretical explanation for it.
Here are two answers using notation:  $H$=top surface of water, $h$=height of jet, $d$=distance traveled by jet


*

*$Pressure = \rho g (H-h)$, Force on a droplet of area $A$ and volume $V$ is $F=\rho g (H-h) A$. Assuming this force acts for some unit time t, speed at orifice exit is $S=g (H-h) At/V$. Since time to fall for the droplet is $\sqrt {2h/g}$.  Distance traveled by a droplet is $d=g (H-h) At/V \sqrt {2h/g}$,  the maximum occurs at $h = H/3$

*Using Bernoulli's equation, $P+\frac{1}{2}\rho S^2+\rho gh=const$, then assuming velocity at top surface is negligible, $\frac{1}{2}\rho S^2=\rho g(H-h)$, so $S=\sqrt {2g(H-h)}$. Since time to fall for a droplet is $\sqrt {2h/g}$.  Distance traveled by a droplet is $d=\sqrt {2g(H-h)} \sqrt {2h/g}$,  the maximum occurs at $h = H/2$. 


In my opinion, the first one is right, as ignoring the velocity of top surface of water is incorrect. Can you help me understand which is the correct approach. 
Note:  Similar question was asked before and they all seemed to take the second approach. Also, I used $S$ for speed, as I used $V$ for volume of droplet
 A: The first approach is incorrect because you identify the time a volume of water gets accelerated by the full force of gravity with the time it would take it to fall that distance. If that were the case, it would see less acceleration because the ram effect would decrease the pressure it sees from above to zero.
The second approach gives you the correct equation for the speed although I find your notion curious. $P_h = P_\mathrm{top} + \rho g (H-h)$ is the pressure at the hole (or rather the difference to the ambient air pressure) and the velocity at which water has to travel or gets accelerated to from zero velocity can then indeed be calculated from Bernoulli's equation because the dynamic pressure $\Delta P = \frac{1}{2} \rho v^2$ also correctly describes the ram effect, the difference in pressure in the forward/backward direction (and that's no coincidence, so identifying this with Bernoulli's law for the opposite change in pressure perpendicular to the flow is fine at least in my eyes). Both of your approaches appear to assume that the jet gets ejected horizontally (which is probably true, but I think it should be stated in case someone making a slanted hole misunderstands). I have not checked your maximization, only that the equations you use for it look correct assuming that $S$ is your speed (others might prefer or expect it to be written as $v=|\vec{v}|$ or $\dot{x}$).
A: 
The experimental observation is that the plot of distance traveled to
  height of hole "appears" parabolic

It's not "parabolic", it's the square root of a parabola (as you have shown below):
$$
d=2\sqrt{Hh-h^2}\;.
$$
This is just semantics, but it is good to get it right; a parabola only has powers 0,1, and 2, e.g., $ax^2+bx+c$, not square roots...

Here are two answers using notation:  $H$=top surface of water,
  $h$=height of jet, $d$=distance traveled by jet
  
  
*
  
*$Pressure = \rho g (H-h)$, Force on a droplet of area $A$ and volume $V$ is $F=\rho g (H-h) A$. Assuming this force acts for some
  unit time t, speed at orifice exit is $S=g (H-h) At/V$. Since time to
  fall for the droplet is $\sqrt {2h/g}$.  Distance traveled by a
  droplet is $d=g (H-h) At/V \sqrt {2h/g}$,  the maximum occurs at $h = H/3$
  
*Using Bernoulli's equation, $P+\frac{1}{2}\rho S^2+\rho gh$, then assuming velocity at top surface is negligible, $\frac{1}{2}\rho S^2=\rho g(H-h)$, so $S=\sqrt {2g(H-h)}$. Since time to fall for a
  droplet is $\sqrt {2h/g}$.  Distance traveled by a droplet is $d=\sqrt{2g(H-h)} \sqrt {2h/g}$,  the maximum occurs at $h = H/2$. 
  
  
  In my opinion, the first one is right, as ignoring the velocity of top
  surface of water is incorrect. Can you help me understand which is the
  correct approach.

Ignoring the velocity at the top surface of water is correct if the size of the poked holes is much smaller than the cross-sectional area of the water at the top. Additionally, as shown below, ignoring the velocity of the top surface of the water does not affect your calculation of the best height $h=H/2$. (As an aside, you are also ignoring a bunch of other thing that don't seem to bother you. For example, you are also ignoring the difference in atmospheric pressure at the height of the top surface H and the atmospheric pressure outside at the hole at h...) 
Where in your first "answer" did you not ignore the velocity at the top?  Had you not ignored the velocity of the water at the top there would be some location in your derivation where the ratio of the top surface area to the hole area entered the equation... which it does not. You still ignored the top velocity, you just have done a botched job of re-deriving the Bernoulli equation, so you get the wrong answer.
The Bernoulli equation reads:
$$
\rho g z_1 + \frac{1}{2}\rho g v_1^2 + P_1 = \rho g z_2 + \frac{1}{2}\rho g v_2^2 + P_2\;,
$$
and ignoring the velocity of the top surface amounts to setting:
$z_1=H$,$v_1=0$,$P_1=P_{atm}$,$z_2=h$,$v_2=S$,$P_2=P_{atm}$, thus
$$
\rho h H=\rho g h + \frac{1}{2}\rho S^2
$$
Not ignoring the velocity at the top surface amounts to setting:
$z_1=H$,$v_1=S_{top}$,$P_1=P_{atm}$,$z_2=h$,$v_2=S$,$P_2=P_{atm}$, thus
$$
\rho h H+\frac{1}{2}\rho S_{top}^2=\rho g h + \frac{1}{2}\rho S^2\;,
$$
and for an incompressible fluid
$$
S_{top}=S\frac{a_{hole}}{a_{top}}\;.
$$
So, if you wanted to you could take into account the velocity at the top by replacing 
$$
S^2 \to S^2(1-\frac{a_{hold}^2}{a_{top}^2})\;,
$$
but you would have to know the area of the hole and the area of the top surface.
Furthermore, the above correction clearly does not change the shape of the $h$ dependence, so the maximum still appears at $h=H/2$. I.e., the range is now
$$
d=2\sqrt{\frac{\rho(Hh-h^2)}{1-\frac{a_{hole}^2}{a_{top}^2}}}\;,
$$
which is still maximized by $h=H/2$
A: I actually tried this, for an introductory physics class, punching holes in a gallon plastic container in which the area of top of the container, A, was about 23,000 mm^2 and the area of the small hole,A', was about 1 mm^2  the equation of continuity, i.e. Av=A'v' says that the velocity of the water at top surface,v ,is almost negligible
compared to the velocity coming out the small hole,v'. 
The Bernoulli equation seemed to be the first choice (however as my experiment showed
and I will comment on below, possibly the Navier Stokes might have been a better choice)
The Bernoulli equation, with the unprimed quantities at the top water surface and the primed quantities at the small exit hole, is.
                          **P+(1/2)ρv^2+ρgh=P'+(1/2)ρv'^2+ρgh'**

If there is no vorticity, the water swirling around the drain effect, and since there is little difference in atmospheric pressure P=P' and v=0 and if the height is measured from h' then the velocity out of the hole should be √2gh.
this velocity can be checked if the exit hole is a height H from the floor and the stream hits a horizontal distance, x, from the hole projection is given
by the equation
                                      **v'=x√g/√2H**

When I actually did this experiment the directly measured horizontal velocity was only about a half of the Bernoulli calculated velocity.  I did notice that the water seemed to be swirling i.e. vorticity, suggesting that a Navier Stokes treatment might explain this.
I found out that there is no general solution to the Navier Stokes equation and there was a million dollar prize for solving it. I didn't solve it and decided to do something else.
