If that calculator is correct then the terminal velocity of the water balloon is 208 mph.
Want to get hit by something 4 times heavier than a baseball going twice as fast as the fastest pitcher on Earth?
Thinking some more about this, and trying to avoid getting into the really difficult world of modelling the fluid, there are some high school physics level things we can think about.
Energy is probably not the right quantity to think about. Energy transfer from the balloon to the head of the victim is going to be very difficult to work out. Energy is subject to transfer to a range of forms, so for instance a bullet hitting a helmet may well have most of its energy dissipated as heat in the helmet, and not imparted to the victim as kinetic energy. The ECH most certainly depends upon such a mode. However momentum is much more useful. Conservation of momentum gets us much less equivocal results.
So, some back of the envelope calculations. (Worth exactly what you paid for them. 
)
If we assume that the water drop hits square onto the top of the victim’s head, and the water splashes out in a pretty much horizontal spray, and we assume negligible contribution to the forces from viscosity, friction or the energy needed to rupture the balloon, we can simply look at the momentum change in the drop. This is probably a good assumption for any high speed impact, which are those we are worried about.
It seems that for any sensible balloon size 100m/s is about the terminal velocity (although it requires an unrealistic height to achieve) so that is a limiting case.
We can also assume a cube of water rather than a sphere in order to make a few other simplifications.
So: momentum = mass * velocity. = m * v = p
Impulse = change in momentum * time = p * t
time = velocity / height of cube = v / h = t
mass = height of cube^3 = density of water * h[sup]3[/sup] = 1000kg/m[sup]3[/sup] * h[sup]3[/sup] = m
Impulse = h[sup]3[/sup] * v * 1000 * (v/h)
= h[sup]2[/sup] * v[sup]2[/sup] * 1000
So, a limiting case. A 1 litre cube of water falling at 100ms[sup]-1[/sup]. This has a momentum of 100kgms[sup]-1[/sup] and takes 0.001s to travel its own length. Thus the force imparted as it splashes is:
100,000 Newtons for 1 ms.
That will certainly fracture a lot of vertebrae and I would find it hard to believe is in any way survivable.
However we might notice that the terms are not linear, and things get more reasonable fairly quickly.
100ms[sup]-1[/sup] is just silly anyway. So maybe drop it from a 10 story building, call it 30 metres. Final velocity is 24ms[sup]-1[/sup]. Close enough to be one quarter terminal velocity, and so our force drops 16 fold to about 6,200 Newtons over 4 ms. That is still a very serious force, and would likely injure someone, possibly badly.
Final velocity goes up as the square root of the height of the building, whereas the force of the balloon goes up as the square of the velocity, so usefully the force goes up linearly with the height of the drop. So, for a 1 litre balloon about 500 Newtons per metre of drop height. Note, for low drop heights most of the above assumptions don’t hold, and the force will be substantially less. From a height of 1 metre the balloon does not explode in a horizontal spray of water, indeed the balloon might not even rupture, and if it does, the water runs down over the victim, transferring momentum over a long period of time.
The force falls slower than the mass falls, because the height of the cube of water falls, and that means the momentum changes over a shorter time.
We can also look at the velocity of the victim. Assume a 100kg victim. (A bit chubby, but it makes things easier.) Hit by a 100kgms[sup]-1[/sup] balloon, if we assume they are a rigid body, they will end up falling at 1 ms[sup]-1[/sup]. Just imparted to their head, (assume 5kg mass) this momentum would result in their head being accelerated to 20ms[sup]-1[/sup] (44 miles/hour) onto their neck. I think we can assume this will break their neck. Our ten story building drop gets us a head travelling at 11 miles an hour, which is likely to cause significant injury, but probably not kill them. The momentum goes up as the square root of the height (ignoring atmospheric drag), so even a 5 story drop is likely to be within the realm of causing neck injury. Below about a 5 story drop I susepct the main assumptions start to fail, and the forces start to fall much quicker.