What is the coefficient of friction between an inclined plane and a body moving along it, if the body is launched

What is the coefficient of friction between an inclined plane and a body moving along it, if the body is launched up the inclined plane with an initial speed V0 = 6m / s and stopped after passing a path of 1.6 m, the angle of inclination of the plane to the horizon a = 45.

Vo = 6 m / s.
V = 0 m / s.
S = 1.6 m.
α = 45.
g = 9.8 m / s ^ 2.
μ -?
Changing the total energy went to overcome the work of the friction force: ΔE = A.
ΔЕ = m * Vo ^ 2/2 – m * g * h, where m is the mass of the body, g is the acceleration of gravity, h is the height to which the body has risen.
h = S * sinα.
ΔE = m * Vo ^ 2/2 – m * g * S * sinα.
A = Ftr * S.
Ffr = μ * N, where μ is the coefficient of friction, N is the reaction force of the support.
N = m * g * cosα.
m * Vо ^ 2/2 – m * g * S * sinα = μ * m * g * cosα.
Vo ^ 2/2 – g * S * sinα = μ * g * cosα.
μ = (Vo ^ 2/2 – g * S * sinα) / g * cosα.
μ = ((6 m / s) ^ 2/2 – 9.8 m / s ^ 2 * 1.6 m * √2 / 2) / 9.8 m / s ^ 2 * √2 / 2 = 1.
Answer: μ = 1.