A ball is projected with kinetic energy E , at an angle of 60 ° to the horizontal. The kinetic energy…

$E=\frac{1}{2}m{u}^2$

At the Highest point, velocity is equal to the horizontal component of the velocity of projection. Therefore,

$v=u\cos 60°=\frac{u}{2}$

$\therefore$ Kinetic energy at the topmost point $=\frac{1}{2} m{\left(\frac{u}{2}\right)}^2=\frac{E}{4}$