A shot 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.