It was already discussed in the comments that a water rocket needs to push "something" out. It is instructive to do the calculation in a little more detail to see where the "energy" goes. For this I will consider the relative share of energy going to the rocket and the "expelled matter" (gas, or water) as a function of the expelled mass. To simplify things, we will assume that all matter is expelled as a single entity with a certain velocity; in reality you might need to integrate, but any inequality that holds for a small amount of expelled matter will hold for the integral over many such amounts.
I will use upper case symbols for quantities relating to the "rest of the" rocket (mass M, velocity V, momentum P - without the expelled mass) and lower case for the expelled matter(m, v, p). From conservation of momentum, $P = -p$ so $M\cdot V = - m\cdot v$. The energy of the rocket $E_r$ and expelled mass $E_m$ will be respectively:
$$E_{r} = \frac12 M V^2 = \frac{P^2}{2M}\\
E_m = \frac12 m v^2 = \frac{p^2}{2m} = \frac{P^2}{2m}$$
It follows that the ratio of (energy in rocket)/(energy in expelled matter) is
$$\frac{E_r}{E_m} = \frac{m}{M}$$
In other words - the lower the mass of the expelled matter, the greater the relative amount of energy it contains. In the limit of "no water", the little bit of air mass contains virtually all the energy.
I apologize in advance, but I'm going to make the problem look worse first so that I can explain it:
Imagine that instead of you measuring the velocity from the station where the rocket is launched - let's call it station A - you measure instead from station B, which is moving away from station A at some high speed. And imagine that station A launches two identical rockets, one toward you and one away from you.
Since we're operating under the well-tested current theory of special relativity, we understand that we can either say "B is moving away from A" or with equal correctness "A is moving away from B" since we can declare either one as "stationary" for purposes of measuring velocities with respect to our chosen coordinate system. So let's start over and say that Station A is moving at speed away from Station B and launches two rockets.
Of course, before they launch from A, both rockets will have a certain kinetic energy based on their motion, and that energy will be equal for the two identical rockets. When we launch them, however, that begins to change: one of them is starting to move away from us even faster, while the other begins moving away from us slower. After some time, in fact, one will be moving away from us at twice the speed it originally had - that is, four times the kinetic energy - while at the same time the other is moving away from us at zero relative velocity: no kinetic energy whatsoever! Now we're really mixed up, because both rockets did the exact same thing but one ended up with a very high kinetic energy and one with no energy whatsoever.
The issue I hope to highlight with that example is that kinetic energy is frame-dependent and not an intrinsic energy for an object. That's why you can safely ride along in a car moving at 100km/h along a road but not be safely struck by a car moving at 100km/h along a road; if you travel with the car, it has zero kinetic energy in your frame, but if you're standing on the road it has a very high kinetic energy in your frame. Kinetic energy is relative.
On another note, I said before that the two rockets are doing "the same thing," but in any given single frame of reference that's not quite true. Obviously, in my example above, from the frame of station B one rocket is losing kinetic energy and the other is gaining kinetic energy, for example. But that's only considering the rockets themselves: those are being propelled by launching photons the other direction, and those photons are expelled with a certain energy themselves. If you were to measure the energy of the photons themselves, (attaching a small mirror to the back of the rocket headed toward you so that you can see the "exhaust") you would notice that over time the exhaust of the rocket moving away from you seems less energetic and the exhaust of the rocket moving toward you seems more energetic. That is to say, photons launched toward you are less energetic than photons launched away from you... which was the same conclusion we came to with the rockets themselves launching from Station A, and again is a consequence of the relativity of kinetic energy. But this time we can more directly say that one is "red-shifted" and the other "blue-shifted."
EDIT: It may also be simpler to imagine the solution if you use a normal reaction drive with massive exhaust. The exhaust is low mass but leaves at high speed; the rocket is higher mass and thus gains less velocity from the reaction. The kinetic energy increases by the same amount in opposite directions for each body of mass (taking all the exhaust as a "body"), otherwise we're getting energy for free from somewhere. Also, since massive exhaust has subliminal expulsion speed, you can have a case where the rocket is accelerating away from you but the exhaust is also moving away from you, which isn't possible with photon drives. But in all cases, photons or not, the total energy of the system is conserved: measured from an inertial coordinate system, the rocket "gains" the same amount of energy that the exhaust "loses," and the measured exchange rate does increase quadratically.
Best Answer
If the pendulum is in equilibrium then the rocket motor does no work on the pendulum. It exerts a force on the pendulum, but because the pendulum is not moving, this force does no work on the pendulum. It is exactly as if the pendulum was held by a length of rope - the rope exerts force on the pendulum but does no work on it.
The rocket motor, of course, does work by expelling its exhaust, but the energy that goes into the exhaust is initially seen as kinetic energy of the exhaust, and is eventually dissipated into the environment as sound and heat.
Note that during the initial phase of the motion - as the rocket motor moves the pendulum from vertical to its new equilibrium position - the velocity of the exhaust is lower than in the equilibrium position. This is because the exhaust has a fixed velocity relative to the rocket motor, which is now moving. So in this initial phase the rocket motor does less work on the exhaust and does some work on the pendulum instead - and this energy is stored as the potential energy of the pendulum in its new equilibrium position.