Alas, the method proposed does not work. The Earth's "terminator" is a broad band, the penumbra of the Earth's shadow. For the case in question, the terminator would be a band hundreds of feet wide (at the top "edge" of the penumbra, the full disk of the Sun is visible resting on the horizon, at the lower "edge" the full disk has just barely disappeared below the horizon). You would not be able to pin down the location of the terminator to better than a few hundred feet, and so of course you wouldn't be able to measure its acceleration. After consider, have you ever seen even a photo of a building with a shadow on it like this? Where the shadow is coming from the horizon and not some nearby structure?
So what can you do instead...
1) You can attempt to time the exact instant of sunset from two different heights (as described earlier today in the post forwarded from "rete"). This is a trick that has been known for centuries.
2) You could also measure the variation of the "dip" of the horizon with height. The dip is given by alpha*sqrt(2*h/R) where R is the radius of the Earth and alpha is a factor, near one, which depends on terrestrial refraction in a relatively constant way (and can be measured independently). So you get out a sextant and you measure the altitude of the Sun from two different heights at the same instant. One observer gets h0+alpha*sqrt(2*h1/R). The other gets h0+alpha*sqrt(2*h2/R). The difference between them is alpha*[sqrt(2*h1/R)-sqrt(2*h2/R)]. Plug in a good estimate for a and then solve for R.
3) You measure the dip directly. This can be done with a carefully plumbed theodolite or similar instrument. You just measure the angle down from the zenith to the horizon and use the dip formula again. You can also do this with two observers using hand-held sextants by measuring the altitude of a celestial body within 20 or 30 degrees of the zenith, one measuring to the horizon directly beneath the object and the other measuring to the horizon on the opposite side of the sky (there is a procedure called "swinging the arc" that guarantees you measure the angle directly to the horizon). They measure these altitudes simultaneously from the same height. The excess over 180 degrees divided by two gives the dip. Then use the dip formula again.
4) You can use the maximum distance to the horizon seen from different heights. Without refraction, this distance, expressed in nautical miles, is equal to the geometric dip in minutes of arc. Someone mentioned Classical Greece... While it's hard to see small vessels disappearing over the horizon without optical aid, it was well-known to all seafaring peoples that you can see much farther from the masthead and farther still from a lookout point on a mountain on the coast (and the Greeks have lots of these!). You can see roughly ten miles from an altitude of 100 feet and roughly 30 miles from a thousand feet up. This method has low sensitivity since objects appear to sit right on the horizon over a broad range of distances. Expect no better than 50% accuracy in a determination of the Earth's radius from a method like this.
Two more thoughts:
First, it's a relatively easy problem in plane geometry to determine the formula for "geometric dip" which is sqrt(2h/R). The only tool needed beyond a right triangle is the small angle approximation for the cosine. So that's a good "high school" math problem. It's a much more interesting problem, involving real physics and some interesting and rather messy physical details, to determine the realistic dip formula and calculate that factor alpha. That's a good college level physics problem though most of the messy details would probably be dismissed by most physics folks as "weather". :-) You can show that if temperature in the air rises rapidly with altitude (as it sometimes does), then the realistic refraction formula gives zero dip and the Earth appears to be a flat plane for all optical observations.
On another point, that penumbra of shadows cast by the Sun can be put to good use. The angular diameter of the Sun is just over half a degree or an angle of 1/108 when expressed as a pure angle (a.k.a. radians). Equivalently, the ratio of the Sun's distance to its diameter is about 108. That also is the angular width of the Sun's penumbra. So when you look at a shadow cast by the Sun on the ground from some object of unknown distance, estimate the width of the penumbra and multiply by 108. That yields the distance to the object casting the shadow. Of course, this distance has to be measured on a plane perpendicular to the rays of sunlight but even when it's not, you can usually figure that out quickly. This trick also works with those circular images of the Sun that you see under a canopy of leaves (if the circles are four inches across, the leaves making the images are 36 feet high). And finally, you can determine the extent of the penumbra by looking directly at the Sun (do
n't stare). Walk back and forth along a line perpendicular to the Sun's rays and find the two spots where the Sun is just barely completely uncovered by a distant object and just barely completely covered by a distant object. The distance between those two points is the width of the penumbra at your location. Multiply by 108 to get the distance to the object in question (works with anything that has a vertical edge, otherwise you need to think a little more about the geometry).