Any Physics majors? I can't seem to get the right answer :/ Help me by showing me correct solution? (10 pts!?)

) A 620 g basketball is pushed up a 30° ramp. The ball starts from rest and the length of the surface of the ramp is 5 m. The coefficient of kinetic friction between the surface on the ramp and the ball is 0.1. The force ceases when the ball leaves the top of the ramp and the total horizontal distance traveled by the ball after leaving the ramp is 20 m. What was the magnitude of the pushing force (in N)?

Let v be the velocity when the ball reaches the top of the ramp.
then v cos 30 x t=20

again 0=vsin30 x t -1/2 gt^2
or t=2vsin30/g

putting this value we get=v cos 30 x 2vsin30/g=20


or v^2 sin 60=20xg
or v^2 =20xg/sin60=226
or v =15 m /sec

Now let N be the force applied

then N - 0.1xmgcos 30 is net force on the ball.

So 2(N - 0.1x mg cos30- mgcos 30)x5/m=15^2

or N= 15^2m/5x2 + 1.1mgcos30

or N= 13.95 + 5.78
N=19.73 Newton.

pushing force F = m g [ sin(theta) + uk cos(theta)] = 0.62 x9.8 ( 0.5 + 0.1 x0.866) = 3.564 N.