For a given volume, light things float and heavy things sink. The cup sinks when you fill it with water because it becomes heavier, and therefore more dense. When the cup becomes more dense than water, it sinks.
The cup would sink just as well if you filled it with rocks, lead, etc. The condition for the cup to sink is that its weight must be greater than the weight of the water it displaces (i.e. its weight must be greater than the weight of a cup exactly the same size, but made out of water and filled with water.)
You can read a number of other questions on similar material here: https://physics.stackexchange.com/questions/tagged/buoyancy
If you take a pin and place it into water very carefully with the pin pointed down, no matter how careful you are, the pin will sink. However, you can easily do the experiment at home where you carefully lay a pin on its side on the surface of the water and presto, it floats. The volume doesn't change, but the surface area in contact with the water does.
The equation relevant to this situation is as follows:
$$F_W=2\gamma L\cos\theta$$
where $F_W$ is the weight of what you want to float, $\gamma$ is the surface tension, $L$ is the length of the object (let's use the pin example), and $\theta$ is the angle of contact with the surface (which changes to make the equation work, if possible). IF this equation is satisfied, the object will float on the surface. Given that $\gamma$ is a fixed value and $\cos\theta$ can only vary between $0$ and $1$ (negative numbers wouldn't make any sense whatsoever in this case), it's pretty obvious that there's a minimum value of $L$ for which this equation will hold.
For our pin of mass $m$, the minimum length contacting the surface of the water would be:
$$L\ge\frac{1}{2}\frac{mg}{\gamma}$$
This corresponds to $\cos\theta=1$. If $L$ is any smaller than this value, then the equation at the top cannot be satisfied no matter what and the pin sinks. From this, it's plain to see that when you try to float the pin on its tip, the length contacting the surface, $L$, is way too small. But it turns out that on its side, $L$ is long enough that some angle $\theta$ allows this equation to be satisfied. Thus, the pin can float.
If we go back to the coin flattened by a hammer, before flattening, the case likely is that $L$ is too small. However, the coin can be flattened to increase $L$ enough that it satisfies the inequality above, and thus, can float on water.
Best Answer
If you are above the water you will get accelerated down until the weight of the water you disperse is equal to your own weight (calling this level $x$). As soon as you are completely submerged the gravitational force downwards will be $\rho Vg$ and by Archimedes principle the force upwards will be $\rho_w V g$, where $\rho$ is your density, $V$ is your volume and $\rho_w$ is the density of water. Thus the total downwards force is $Vg(\rho - \rho_w)$.
So depending on your density you will keep accelerating down ($x$ is never reached), or start accelerating up as soon as you pass level $x$. As humans are almost $90\%$ water, our density is very close to that of water (most people are a little more dense). Such that you probably can push your density to be higher than water if you complete exhale, then you will accelerate right to the bottom of the pool.
If you are less dense you will start accelerating up as soon as you reach level $x$ this acceleration plus the drag (which would depend on your position) will slow you down until your velocity is zero, then you will get accelerated upwards, and if you can hold your breath long enough you should oscillate around level $x$ with an exponentially decaying amplitude. And depending on your initial velocity at point $x$ and your density, the amplitude will vary.
It is basically a dapped harmonic oscillator, but not quite because the force for when you are higher then level $x$ is not linear in position.