A skiier starts at a height of 100m and goes down a slope before going
through a circular loop (like that of a rollercoaster) of radius r.
Considering that the height of the loop is 2r, there is no friction,
and energy is conserved, what is the maximum radius, r, of the loop?

Assume gravity as being 10m/s^2.

I have two methods to solve, and I'm not sure if either are correct.

METHOD ONE: Conservation of Energy without Friction. 
Since there is no friction, an object that falls 100m is able to come
back up to a height of 100m.

- Find velocity due to going down a slope of 100m:
PEi + KEi = PEf + KEf
(mass * 10m/s^2 * 100m) + 0 = 0 + (0.5 * mass * vf^2)
mass cancels out
1000 = 0.5 * vf^2
vf = 44.72m/s

At the top of the loop, KE is minimized and PE is maximized:
PEi + KEi = PEf + KEf
0 + 0.5 * m * 44.72^2 = m * 10m/s^2 * 2r + 0
1000 = 20r
r = 50m

METHOD TWO: Using rotational motion
We assume that the rotational acceleration must remain at 10m/s^2 to beat gravity.

a = v^2 / r
v^2 = 10r
v = sqrt(10r)

Plug back in to equation:
KEi + PEi = KEf + PEf
0.5 * m * 44.72^2 + 0 = 0.5 * m * (sqrt(10r))^2 + m * 10 * 2r
1000 = 0.5 * 10r + 20r
1000 = 25r
r = 40m

I don't see the reasoning behind setting a = 10m/s^2 in the second
method, but that was provided by my peers. So, is either solutions
correct?