Ashot is fired at an angle of 60 degree horizontal with kinetic energy e. if air resistance is ignored, the k. e at the top of trajectory is:
0, e/8, e/4, e/2

I'm not sure what "60 degree horizontal" means.

I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith. 

Now, I'll answer the question that I have invented.

When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is    S cos(60)  =  0.5 S ,
and the vertical component is   S sin(60) = S√3/2  =  0.866 S .  (rounded)

-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.

-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change. 

-- So at the top of its trajectory, its KE is 0.25 of what it had originally. 

That's  E/4 .