Physics frictionless pipe

A child slides down the outside of a smooth frictionless pipe 20 m in radius. How fast is she going when she leaves the pipe?

The answer is 11.55 m/s, but can someone help me understand?

Force balance when she leaves the pipe,
∑F = 0
Fs - mg cos θ = 0
Fs = mg cos θ, Fs = mV^2/R
mV^2/R = mg cos θ
cos θ = V^2/gR, V^2 = 2gh
cos θ = 2gh / gR
cos θ =2h/R

Define cosine angle,
cos θ = (R - h) / R
(2h) / R = (R - h) / R
2h = R - h
3h = R
h / R = 1 / 3
h = R/3, She leaves at h = (20/3) m

V = [2gh]^0.5
V = [(2*10*20)/3]^0.5
V = 11.547 m/s

you need the length and angle of the pipe she slides on.

A drawing would help, I think your use of "pipe" is different from mine.