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 $h=H/2$
