A particle is projected at 60⁰ to the horizontal with a kinetic energy K. The kinetic energy at the highest point is

initial velocity = u

initial kinetic energy: $K=\frac{1}{2}m{u}^2$

at highest point, vertical component of the velocity will be zero.

so, at height point, velocity will be equal to the horizontal component of the initial velocity.

$v=u\cos {60}^o=\frac{u}{2}$

final kinetic energy: $K_f=\frac{1}{2}m{v}^2=\frac{1}{4}\times \left(\frac{1}{2}m{u}^2\right)=\frac{K}{4}$