I vaguely remember doing a Millikan oil drop experiment in a lab in college, 60+ years ago. Here is an example. With enough measurements the statistical error can become 1%, or as precise as one wants.
So your question is basically about the systematic errors. Most of them would also follow the addition in quadrature rule, since most systematic errors themselves follow a gaussian.
I tried to find laboratory examples from students' efforts but was unsuccessful. I do not see though why the various systematic factors you list could not be within less than 1% at that time. Classical physics measurements were quite sophisticated.
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).
Best Answer
M2, is the paper and M1 is the coin. We are exerting a rightward force $F$ on the paper. It's obvious that the paper moves because of the force we are exerting on it, and it's acceleration is decreased by the friction force between the coin and the paper.(The paper is going rightwards, but the coin is exerting a friction force to the paper, leftwards). Based on Newton's third, because the coin is exerting a force on the paper leftwards, the paper is exerting a rightward force on the coin too. That's the reason the coin will stick to the paper when you're pulling the paper slowly. But the question is what happens when we increase this force. Firstly, the static friction force can not exceed a certain value. Which means if $F$ gets too high, the friction force can not keep up with $F$ and at some point the acceleration of the paper becomes more than the coin. So the coin can not keep up with the paper. So after a while, the coin just slips and falls into the glass.
I want to calculate the minimum force $F$ so that it causes the coin to slip. Imagine the maximum static friction force between the paper and the coin is $f_{k.max}$. So the maximum force that can be exerted on the coin is $f_{k.max}$. Which means the maximum acceleration of the coin is:
$a = F/M1 = \dfrac{f_{k.max}}{M1}$
So if you want to make the coin slip on the paper, you have to make the system move with an acceleration more that $a$. Which means:
$F > a * (M1 + M2) => F > \dfrac{(M1 + M2)f_{k.max}}{M1}$
If the force is below this value, then the coin doesn't slip, because the force exerted on it is not above $f_{k.max}$ but if the force exceeds this value, then $M1$ can not keep up with M2.
Think of the coin and the paper, as a whole system, only when the acceleration exceeds some value, the system breaks.
Forces on the paper:
$\sum{F_{paper}} = F - f_{k} = M_2 * a$
Forces on the coin:
$\sum{F_{coin}} = f_{k} = M_1 * a$
And we know that:
$f_k \leq f_{k.max} => M_1 * a \leq f_{k.max} => a \leq \dfrac{f_{k.max}}{M_1}$
$=> M_2 * a + f_{k} = F, => F \leq M_2 * \dfrac{f_{k.max}}{M_1} + f_{k.max} => F\leq \dfrac{(M_1 + M_2)f_{k.max}}{M_1} $