if gravity is 9.8 m/sec2, how far would you fall in a tenth of a second? Since you have to accelerate from zero it can't possibly be 0.98m?

You would reach a maximum speed of one tenth of 9.8 ( v = at)
in other words 0.98 m/s

Your AVERAGE speed must be half this. ie 0.49 m/s

and in one tenth of a second at this speed you must fall 0.049 m ( about 5 cm s = vt )

Here's a method without any complicated formulae.

Your speed increases by 9.8m/s each second.
So your speed after 0.1s is 0.1 x 9.8 = 0.98m/s

You started at speed = 0, therefore your *average* speed during the 0.1s is:
Average speed = (0 + 0.98)/2 = 0.49m/s

Distance = average speed x time
. . . . . . . .= 0.49 x 0.1
. . . . . . . .= 0.049m

Initial velocity, u = 0 m/s
Acceleration, a = 9.8 m/s²
Time taken, t = 0.1 s
Displacement, s = ? m

s = ut + (1/2)at²
Displacement, s = (0) + (1/2)(9.8)(0.1)² m = 0.049 m

s = u*t + (1/2)*a*t^2
s = distance, u = initial velocity, a = acceleration, t = time.
Initial velocity is zero, so u = 0.
Hence distance = (1/2)*9.8*(1/10)^2 metres
= 4.9/100 metres = 0.049 metres <<<

Δy=Vi*t-1/2g*t^2. Plug in. I think it results in ~1/10 meter, Idk I did it in my head.

s(10) = 1/2 g 10^2
s(9) = 1/ g 9^2
s(10)-s(9)= 1/2 g (10^2-9^2) = 19/2 g=93.1 m
This is the ditance of fall between 9 and 10 s.