A ball is projected with kinetic energy E at an angle of 45? to the horizontal. At the highest point during its flight, its kinetic energy will be

Dear Student
The expression for kinetic energy is KE= 1/2 mv²
When a body is projected with kinetic energy E and angle $\theta$ in a projectile motion, then at its highest point, the body has no vertical velocity ($v\sin \theta =0$) but it has the horizontal velocity ($v_{II}=v\cos \theta$).

Thus, the kinetic energy of the body at its highest point will be,
 $$\begin{gathered} E_h=\frac{1}{2}m{\left(v\cos \theta \right)}^2 \\ {E}_h=\frac{1}{2}m{\left(v \cos 45\right)}^2 =\frac{1}{2} m{v}^2{\left(\frac{1}{\sqrt{2}}\right)}^2 \\ {E}_h=\frac{E}{2} \\ so, at the highest point the kinetic energy is \frac{E}{2} . \end{gathered}$$

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