If I read the description correctly, your experiment is set up to impart an initial velocity to the coin before your timing starts - as you drop it from slightly above the edge of the table, but don't start timing until you hit the edge (hear the impact).
At the moment of impact, the center of mass of the coin will slow down a little bit - but it will still have some residual velocity. An example calculation:
If you drop the coin from distance $h_0$ above the table edge, it will have a certain velocity $v_0$ by the time it hits the edge. If you make a "perfect" hit between the edge of the coin and the edge of the table, the coin will start spinning while continuing to fall; for a coin with mass $m$ and radius $r$, the moment of inertia about the axis is $I = \frac14 m r^2$. The change in linear momentum $\Delta p$ will give rise to a change in angular momentum $\Delta L = r\Delta p$; conservation of energy tells us that the sum of rotational and translational kinetic energy after the impact are the same as before. Finally, the contact force will go to zero when the velocity of the edge of the coin is zero, so $r\omega_1=v_1$. We can now write conservation of energy:
$$\frac12 m v_0^2 = \frac12 m v_1^2 + \frac12 I \omega_1^2$$
which we solve by substituting for $\omega_1$ and $I$ per the above. We are left with the relationship between $v_0$ and $v_1$:
$$v_1 = \sqrt{\frac{4}{5}}v_0$$
so the coin will lose a fraction of its initial velocity when it hits the table. And then it continues to fall distance $h$.
Now a coin that starts with an initial velocity will reach the floor more quickly than a coin starting from rest. The time taken is obtained by solving
$$h = v_1 t + \frac12 g t^2\\
t = \frac{-v_1 +\sqrt{v_1^2 + 2 g h}}{g}$$
Sanity check: when $v_1=0$ this reduces to the usual result $t = \sqrt{\frac{2h}{g}}$, and when $v_1$ is quite small, we can do an expansion of the expression to obtain
$$ t = \sqrt{\frac{2h}{g}} - \frac{v_1}{g}\left(1+\frac{v_1}{4gh}\right)$$
Substituting things in to find the effect on calculated height quickly gets very messy, but it's straightforward to plot the relationship between height above the table from which the coin was dropped, and calculated height difference (where the true value = 45 cm):
Note that the difference in calculated height is a fraction of the height from which you dropped the coin:
\frac{-v+\sqrt{1+\frac{v_1^2}{2gh}}}{g}$$
Also note that the velocity of the coin when hitting the floor (approximately 3 m/s after a 45 cm drop) means that a 20 ms timing difference corresponds to a 6 cm difference in apparent height: that's a massive number.
I conclude that there is another problem with your setup that is not properly described - "something" you are not telling us, or some way in which you are not interpreting your data correctly... In particular I wonder whether your interpretation of the sounds as they relate to the start of the drop are correct. I would like to repeat this experiment using high speed video (most modern cameras and even phone can film 240 fps which should be plenty to see what is causing a 20 ms difference).
I assume you meant "cardboard", not "cupboard".
When the cup is full of water, it is empty of air, and water is relatively incompressible.
So when you turn it over, in order for the water to leak out, the cardboard would have to move a small amount away from the edge of the cup, which it cannot do without expanding the water slightly, which the incompressibility of the water does not allow.
So if the seal around the edge of the cup is good, you cannot move the cardboard without reducing the pressure in the cup, and the air pressure outside is not being reduced, so the air pressure outside holds it in place.
If you did this in a vacuum (ignoring that the water would boil) it would not work. The cardboard would just fall off.
It is essentially no different that pressing the cardboard against a wet plate of glass. where it sticks unless somehow you can inject some air into the space, say by inserting a needle.
Best Answer
In the cases of Bottle A and Bottle C, they are full and empty. So the amount of water in them (or not) can be considered to be a complete system along with the bottle, since there is no possible way in which the fluid in the full bottle would reduce its volume or overall distribution, certain properties like the system's center of mass,center of gravity and its overall mass distribution stays constant.
But considering bottle B, when you try to flip it, the centrifugal force created by the force you apply causes the quantity of water in the bottle to shift, which causes a shift in its center of gravity, and a change in its mass distribution, which can cause an overall change in the bottles rotation.( The moment of inertia of the body changes. The moment of inertia is a property which is analogous to mass, which is used to study rotation. A body with more moment of inertia would rotate less freely and vice versa). So it indeed is because of the quantity of the water in the bottles.