Can you please determine the banking angle needed to make a high speed turn off of a highway exit ramp with a radius of 720 m at a speed of 80 km/h, assuming there is no friction.
I arrived at an answer of: 86 degrees above the horizontal, but that does not seem right.
As the car goes around the banked curve the relevant forces are:
Ff - Force of friction is zero
Fn - perpendicular to and directed away from the road's surface
Fg - the gravitational force straight down
The acceleration will be directed towards the center of the curve - in this case I will assume the this is directed towards the left.
Breaking the normal force into horizontal and vertical components:
Fnv=Fn*cos(alpha) - the vertical component of the normal force
FNh=Fn*sin(alpha) - the horizontal component of the normal force
In the vertical direction there is no acceleration and so opposite force must be equal. In this case that makes the vertical component of the normal force upward equal to the gravitational force downward:
Fg=Fnv becomes m*g=Fn*cos(alpha) 
In the horizontal direction the sum of the forces is equal to the mass multiplied by the acceleration. In this case the sum of the forces is the horizontal component of the normal force and the acceleration is centripetal. Newton's 2nd Law becomes:
SF=m*a which becomes Fnh=m*a then Fn*sin(alpha)=m*v^2/R  since the acceleration is centripetal.
dividing equation  by equation :
Therefore, the angle alpha becomes: alpha=atan(v^2/(R*g))
Solving for the angle alpha: alpha=atan(22.2^2/(720*9.8)=atan(0.0698)=40 degrees