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

1. $Pressure = \rho g (H-h)$, 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 $h = H/3$
2. 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

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

1. $Pressure = \rho g (H-h)$, 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 $h = H/3$
2. 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

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

1. $Pressure = \rho g (H-h)$, 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 $h = H/3$
2. 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.

Note: Similar question was asked before and they all seemed to take the second approach.

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

1. $Pressure = \rho g (H-h)$, 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 $h = H/3$
2. 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