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In calculating the exit velocity of the water in a water rocket, all the sites I have found use Bernoulli's equation in the rockets frame of reference. I know that that is correct if the rocket was an inertial frame eg being held at rest or moving with a constant velocity. However the rocket is being accelerated or put another way is not being held fixed in the inertial frame. I think that makes a differece. If I was standing in the rocket's frame I would feel very heavy. ie I experience a higher g force than normal. So I rather think that the exit velocity of the water is higher than is being calculated but I do not know how to do that. Can anybody tell me I am wrong or direct me to a link that addresses this?

Here is an example link https://www.researchgate.net/publication/243714323_Soda-bottle_water_rockets

Alan Johnstone
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    Can you link to one of the derivations you found? It may be that the derivations are correct if the calculation at a particular time $t$ is being done in the inertial reference frame moving with the rocket at that same time $t$. – Michael Seifert Jan 17 '24 at 13:01
  • I have added the simplest link – Alan Johnstone Jan 17 '24 at 17:08

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One issue here is that the usual Bernoulli equation, that $$ \frac 12 |{\bf v}|^2 +\frac{P}{\rho} +gh $$ is constant along a steamline, does not apply when the velocities of the water is time dependent. Instead, for irrotaional flow where ${\bf v}=\nabla \phi$, Bernoulli becomes $$ \frac{\partial \phi}{\partial t} +\frac 12 |{\bf v}|^2 +\frac{P}{\rho}+gh $$ is constant. Also you are right in that the "$g$" in the above equation will be that felt in the acclerating frame: $g\to g+a_{\rm rocket}$. That extra will tend to "pull" the water out of the rocket in addtion to the "push" from the air.

mike stone
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