A ball whose kinetic energy is $E$, is thrown at an angle of ${45}^{\circ }$ with the horizontal. Its kinetic energy at the highest point of its trajectory will be

The correct option is C

$\frac{E}{2}$

Let $m$ be the mass of the ball. If $v$ is its speed of projection, its kinetic energy is
$E=\frac{1}{2}m{v}^2$

At the highest point, the velocity of the particle is
$v^{\prime }=v\cos \theta =v\cos {45}^{\circ }=\frac{v}{\sqrt{2}}.$ Therefore its kinetic energy at this point is
$E^{\prime }=\frac{1}{2}m{v}^{\prime 2}=\frac{1}{2}m{\left(\frac{v}{\sqrt{2}}\right)}^2{E}^{\prime }=\frac{1}{2}\times \frac{1}{2}m{v}^2=\frac{E}{2}$

Hence, the correct choice is ($\text{c}$).