A ball whose K.E is E,  is projected at an angle of 45° to the horizontal, the K.E. of the ball at the highest point of its flight will be,

  $$\begin{gathered} given KE at the point of projection =E=\frac{1}{2}m{u}^2---\left(1\right) \\ at highest point velocity of ball = u \cos \theta \\ KE at the highest point of projectile=\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{2}m{u}^2{\cos }^2\theta \\ given angle of projection \theta ={45}^0 \\ KE at the highest point =\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{2}m{u}^2{\cos }^{2}45 \\ KE=\frac{1}{2}m{u}^2\frac{1}{2}---\left(2\right) \\ substitute eqn \left(1\right) in eqn\left(2\right) \\ KE at highest point=\frac{E}{2} \end{gathered}$$