What is the maximal height for a water rocket's flight?

Question

A water rocket works like this: there is a circular slot of area $A_1$ at the bottom centre of a cylinder of cross-sectional area $A_0$ and height $L$ that is filled with water to an initial height $h_0$. This slot will fall away during launch. The water has pushed all of the air that was originally in the cylinder to the top $L-h_0$ of the cylinder (I believe this is an isothermal compression: the compression was fast), so is at a higher pressure of $\frac{L-h_0}{L}P_0$, where $P_0$ is atmospheric pressure.

To launch, the slot is instantaneously removed (leaving a hole of area $A_1$ in the bottom of the cylinder), and water is pushed downwards, as the air pressure is higher inside the cylinder than outside, at a speed $u(t)$. There is no sloshing of the water in the cylinder: the body of water remains cylindrical. Thus the air in the cylinder now takes up more volume (has expanded in an adiabatic expansion), but because of the upwards impulse imparted to the cylinder by the leaving water, the cylinder is now moving upwards with speed $v(t)$. The rocket will reach a maximum height $H_{max}(h_0)$, where $h_0$ is the original height of the water. What $h_0$ will give the maximum value of $H_{max}(h_0)$ for fixed $A_0,A_1,l$?

Partial solution.

In the adiabatic expansion, let $V(t)$ be the volume of the air in the rocket and $P(t)$ the pressure. As the air is mostly diatomic (Nitrogen and Oxygen are),

$$\displaystyle P(t)V(T)^{\frac{1+\frac{5}{2}}{\frac{5}{2}}}=k$$ $$P(t)V(t)^{\frac{7}{5}}=A_0P_0 \frac{(L-h_0)^2}{L}$$

$$ \frac{dV}{dt}= A_1 u(t)$$

The change of momentum per unit time of the water being spewed out the bottom is

$$\rho \delta V(v(t)+u(t))$$

Answer 1

Useful facts:

1. The non-relativistic rocket equation, which governs the trajectory of a rocket, is \begin{align} \mathbf F = M\frac{d\mathbf v}{dt} - \mathbf u\frac{dM}{dt} \end{align} where $\mathbf F$ is the next external force on the rocket, $M = M(t)$ is the mass of the rocket plus it's fuel at time $t$, $\mathbf v$ is the velocity of the rocket as measured from some inertial frame (the Earth frame can probably be taken approximately inertial for your purposes here), and $\mathbf u$ is the exhaust velocity relative to the rocket.

2. If the rocket is launched in the gravitational field of the Earth, then $\mathbf F = M\mathbf g +\mathbf F_\mathrm{air}$ where $\mathbf g$ is the acceleration due to gravity and $\mathbf F_\mathrm{air}$ is air resistance.

Take special note of the second term on the right hand side of the rocket equation where the rate of change of the mass of the rocket plus its fuel appears. This shows that the motion of the rocket depends crucially on the rate at which the fuel mass is ejected.

One you know the speed of the ejected fuel relative to the rocket and the rate at which the fuel mass is ejected, you can attempt to use the rocket equation to solve for its trajectory.

In the special case that (a) we neglect air resistance, namely when $\mathbf F_\mathrm{air} = 0$ (b) we assume that the launch takes place suffciently close to the Earth's surface to approximate $\mathbf g$ as a constant and (c) the exhaust velocity $\mathbf u$ is constant, the general solution to the rocket equation is \begin{align} \mathbf v(t) = \mathbf v_0 + \mathbf u \ln\frac{M(t)}{M(0)} - gt \end{align} In particular, the final velocity of the rocket after it has used all of its fuel is \begin{align} \mathbf v_\mathrm{final} = \mathbf v_0 + \mathbf u \ln\frac{M_\mathrm{final}}{M(0)} - gt_\mathrm{final} \end{align} To maximize the height, we want to maximize the final velocity. This means you want to minimize the ratio $M_\mathrm{final}/M(0)$ and $t_\mathrm{final}$. This means you want as much as possible of the initial mass of the rocket to be comprised of fueld, and you want the burn time to be as short as possible. This is closely related to why rocket launches are usually so explosive.