Why does the falling of water droplets look the same as an expanded balloon released? 
Today I was observing water droplets and I noticed that the drop falling out almost looks like the same if I blow into a balloon with my mouth and let it go. I mostly understand the mechanism of why the drop let stays and falls: due to surface tension initially a hemispherical type surface is formed which then falls when the weight amassed is much greater than the mass of the weight surface tension could support.
On the other hand, consider the balloon, we blow into it increasing it's volume and when we release it, it propels forward by ejecting mass similar to a rocket. Why exactly does this ballon phenomena look so similar to how the droplet behaves?
 A: As the droplet leaves the tap, it forms a bulge at
the mouth of the tap (that is held together by surface tension). As more water enters the droplet, it begins to fall while forming this "pear shape" analogous to a "balloon" as you have called it.
This happens because the fluid at the bottom of the droplet begins to fall at a faster rate than that of the water entering the drop from above. This will cause the "pear" to
develop the thin neck (like a balloon) as you can see in the diagram below. Because gravity is pulling the drop down, the neck then becomes thinner, until it finally breaks so that the drop falls.

The phenomena described above is similar to a balloon, in that there are tension forces working on both objects. There is tension in the material making up the balloon, and this tension increases as the balloon is inflated (increased pressure), or $$T=\frac{R\Delta p}{4}$$ where $\Delta p$ is the difference between the inner and outer pressure.
But the surface of the droplet also has a tension, called surface tension. The surface tension is also related to pressure by the same equation (but with a $2$ in the denominator). When the droplet above increases in size, at some point the droplet gets large enough that gravity will cause the neck of the pear to thin out and break, as explained above.
The reason why the droplet example looks similar to a balloon, is because the material comprising the balloon is thicker (stronger) at the region close to the neck of the balloon, which is why it takes that pear form when it is inflated.
If we release the balloon, the tension (pressure) will squeeze out the air, forcing it to fly in whatever direction we held it. But in this case, the balloon flies away due to propulsion, whereas for the drop it was due to gravity and surface tension both working to deform and accelerate the drop.
A: A water drop is a kind of balloon, in that it is a fluid enclosed in a flexible surface which is in tension. In the case of the balloon the surface in tension is the rubber. In the case of the water droplet the surface in tension is the surface of the water itself. The tension provides a force which pulls the surface in, but the water drop is hard to compress so its volume is fixed (for a given amount of water). So the tension acts to give to the water the minimum surface area consistent with its fixed volume. When the water leaves the tap this minimum area is a sphere, but the sphere oscillates a bit because the water is in motion from its release from the tap. Then as it falls the sphere gets a bit distorted by air resistance.
Before the drop is released from the tap, the surface of the water reaches to the metal of the tap, so that part is fixed. In this case the surface tension still acts to minimise the surface area of the drop as it forms, but it has this constraint that the top has to be so as to match the tap, and there is also the need to provide a vertical force to support the weight of the drop. The result is that shape you noticed.
The water drop here does just what a thin rubber membrane would do if filled with some heavy liquid, so again the comparison to a balloon is very close.
