This is a common class of question concerning projectile motion and the independence of orthogonal forces.

First the student should sketch the problem an label the forces, In doing this they will see that the only force acting upon the projectile is gravity.

Next the student should recognise that at the maximum height point the ball changes directing meaning that its velocity at that point passes through zero.

As the forward motion V_{x} of the ball is unaffected the kinetic energy of the ball at this point is simply 1/2 m V_{x}^{2 }.

The student now has to calculate V_{x } from the initial velocity V using trigonometry. V_{x }=cos(30)*V (using the classic soh cah toa)Dividing the final energy by the energy at the apex and simplifying gives ( 1/2 m V _{x}^{2})/(1/2 m V^{2}) =V_{x}^{2 }/V^{2 }= V^{2 }cos^{2}(30)/V^{2 }=0.75 so the answer is that the final energy is is 0.75*K