Suppose you have a glass of water, and you pour on it a volume V of water of the same density; if we consider the archimedes principle, what happens to the water that is poured? Does it float, sink, not move?
Archimedes Principle will not cause the poured water (at the same density) to float or sink. Generally however, the incoming water will sink due to its downward momentum, in proportion to the pouring height above the water surface, easily overcoming the surface tension of the incumbent water.
Any air bubbles trapped in the incoming water will then rise to the surface, along with some water surrounding it. See the picture below.
If you would like to view the interaction of the two bodies of water, you could colour one of them with a dye that insignificantly varies the water density, and then conduct the pour testing. Alternating which one (dyed or not) you pour should allow you to draw a confident conclusion.
Archimedes law expresses relates the difference between actual weight (ideally weight in vacuum) to apparent immersed weight (weight in the medium) to density. This is through the following formulations:
Apparent immersed weight = Weight of object - weight of displaced fluid ...(1)
Weight of object/ weight of displayed fluid = Density of object/ density of fluid ...(2)
Since the object here is water with the same density, the above equations show that there is no apparent immersed weight. This means that the buoyant force (on the added water) is exactly equal to the weight of the (added) water, thus the added water will remain suspended wherever you add it.
Now, where you add it is a bit complex since it necessarily involves fluid dynamics, not just statics. Some water will sink due to momentum, as mentioned in the previous answer. Some will splash. Depending on the height of the splash, it could sink or remain at the surface. Once all these dynamic situations cease, each drop/unit of water will remain where it is.