A body is projected with kinetic energy K at an angle of 60° with the horizontal. Its kinetic energy at the highest point of its trajectory will be

At the highest point, the velocity has only the horizontal component vx=v cos⁡θ=v cos⁡60∘=v/2.Now kinetic energy 12mv2 is proportional to v2. Since the velocity is reduced to half, the kinetic energy becomes one-fourth, i.e., K/4.