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).
