Why doesn't the ball have rotational energy after it leaves the ramp? I am having trouble solving #13 from the 2010 F=MA contest:

A ball of mass $M$ and radius $R$ has a moment of inertia of $I = \frac{2}{5}MR^2$. The ball is released from rest and rolls down the ramp with no frictional loss of energy. The ball is projected vertically upward off a ramp as shown in the diagram, reaching a maximum height $y_\text{max}$ above the point where it leaves the ramp. Determine the maximum height of the projectile $y_\text{max}$ in terms of $h$.


From $mgh=\frac{1}{2}mv^2+\frac{1}{2}Iw^2$ and the condition for rolling friction $v=Rw$ it is easy to find the speed $v$ of the center of mass of the ball when it reaches the flat part of the ramp.But we need to know something about the rotation of the ball when it reaches $y_{max}$ in order to use conservation of energy again. In the official solution it is assumed that the ball has no rotational energy after being projected off the ramp, but I don't see why this is true.
 A: Think about this: how does the ball's rotational speed change between the time it is runs off the end of the ramp and the time it reaches $y_\text{max}$? (What torques are being exerted on it between those times that could change its rotational speed?) When it reaches $y_\text{max}$, what kinds of energy does it have? If you can figure that out then you should be able to see why they solve the problem the way they do.
A: Let $B$ denote the vertical position of the lower dashes. Conservation of mechanical energy gives 
$$\tag{1}Mgh + 0 + 0  = 0+\frac{1}{2}M v_B^2 + \frac{1}{2}I\omega_B^2,$$
where the zeros on the LHS represent the fact that linear kinetic energy is zero and rotational energy is zero at the starting point. The zero on the RHS represents the fact that I have defined the potential energy to be zero at $B.$  
Notice that as long as there is rolling, there is also the relation (for circular objects) 
$$\tag{2}\frac{v}{R} = \omega,$$ (an easy way to remember this relation is to consider the dimensions of the quantities involved: $\omega$ "is per second" hence we need to "divide out the length" in $v$ by putting the $R$ in the denominator). 
Now, think about what happens to the ball once it reaches and moves beyond $B$. Should the ball speed up/down? Should it all of a sudden stop rotating? 
You say that the solution says that as the ball reaches the highest point (on the right) it has no rotational kinetic energy? Ask yourself if this makes sense. What happens to the rotational energy? Also you might ask yourself what the velocity of the ball should be at the highest point. 
A: Let us perform conservation of energy for the ramp;
$mgh=\frac{1}{2}mv^2+\frac{1}{2}Iω^2$ 
given the no-slip condition for rolling friction, $ω=R/v_{cm}$ and moment of inertia $I=\frac{2}{5}MR^2$
$mgh=\frac{1}{2}mv^2+\frac{1}{5}mv^2$ 
$v_{cm}^2=\frac{10gh}{7}$ 
Nowhere in the alluded-to solutions is it stated that the rotational kinetic energy is zero upon leaving the ramp. During this period, rotational kinetic energy remains constant due to lack of torque, whilst translational kinetic energy decreases until 0. Thus, only initial translational kinetic energy and final gravitational potential energy remain in this subsequent application of energy conservation.
$mgy_{max}=\frac{1}{2}mv_{cm}^2$
$y_{max}=\frac{5}{7}h$
A: As you mentioned that the friction is absent, ball(it's a solid spherical ball, moment of inertia = (2/5)m*r^2) will slip all the way and will not be able to roll, hence will not rotate, only translatory motion will account for. 
If you had friction on the track, Energy would surely have been distributed between kinectic translational and rotational.
