Now, if we start at position d0 = 0, and we move to position d(1.15) = 10.5, what's the initial (and constant) horizontal velocity?
v0 = (d(T) - d0) / T
v0 = (10.5 - 0) / 1.15
v0 = 9.130 m/s
So that's the horizontal velocity. We could write our velocity vector as 9.130 m/s i + 4.289 m/s j, where i and j are unit vectors in the horizontal and vertical direction. The magnitude of the velocity vector, also known as the speed, is simply the square root of the sum of the squares,
s = √(9.130^2 + 4.289^2)
s = 10.087 m/s
So my answer is a little different than yours. Now, note that I'm assuming a value for the acceleration of gravity of 10 m/s^2, and that's the only value we're assuming. (Remember we also assumed no air resistance, and we also have a few implicit assumptions - like that the coin is a particle without size. But these shouldn't be controversial.) If I assume a more accurate value of 9.8 m/s^2, and go back and recalculate the initial vertical velocity, I get 4.174 m/s, and when I calculate the speed with that value, I get 10.039 m/s. So I'd say your answer is in agreement with that.
We can back out the value of gravity to get an initial speed of 9.88 m/s. I get that the acceleration would have to be 6.88 m/s^2, which would put your teacher's coin somewhere other than the surface of Earth!
I hope that helps!