Block projected up a frictionless incline with initial speed V0

[wuffle]

Once again, need help with mechanics, this one I think I did right, but my friend did it somehow different, and I was wondering who of us 2 is right

Homework Statement

A block slides down an inclined plane of slope angle (alpha) with constant velocity. It is then projected up the same plane with an initial speed V0. How far up the incline will it move before coming to rest?Will it slide down again?

Homework Equations

F=ma

The Attempt at a Solution

What I did:Y coordinate
-mg * Cos(alpha)+N=ma(y)=0

X coordinate
ma(x)=-F(friction)+mgSin(alpha)

F(fric)=mgsin(alpha)

WHEN BLOCK SLIDES UP

mgsin(alpha)+F(gravity)=ma(x)
2mgsin(alpha)=ma(x)
a(x)=2gSin(alpha)

S=-V^2/(4gsin(alpha))

What my friend did:

F(fric)=N * μ = mg * cos(α) * μ
F(fric)=mg* sin(α)=mg* cos(α)* μ
μ = tg(α)
S=V0t -0.5at^2
a=μ * mg * sin(α)

t=V0/a

S=V0^2/a -V0^2/(2a),
where a= sin^2(α)/cos(α) * mgOh and I don't really know whether it will slide down or not, I hope you guys can help me!

Thanks in advance.

 

[Spinnor]

Your distance formula looks right but I figured acceleration, a = 2gsinα

You need to know that for a body on a ramp that static friction is greater then kinetic friction. At the instant the moving body stops evaluate the forces on the body and remember the static friction is greater then the kinetic friction.

 

[grzz]

a=μ * mg * sin(α)

where a= sin^2(α)/cos(α) * mg

I do not think the above are correct since the component of the weight is also contributing to the retardation up the plane.