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**Hint**The horizontal velocity at the highest point is calculated by $v\cos \theta $. From the formula of kinetic energy, the kinetic energy of an object is directly proportional to the square of its velocity.

**Complete step by step answer:**

The Kinetic energy can be calculated by the formula –

$E = \dfrac{1}{2}m{v^2} \cdots \left( 1 \right)$

Now, we know the formula of calculating the kinetic energy which is shown in equation $\left( 1 \right)$

Therefore, according to question, it is given that –

The angle made by the ball with the horizontal is ${45^ \circ }$.

So, let the angle made by the ball with the horizontal be $\theta $.

$\therefore \theta = {45^ \circ }$

Let the horizontal velocity of ball ${v_h}$ and vertical velocity of the ball be ${v_v}$

When the ball is at the highest point, the vertical velocity of the ball becomes zero.

$\therefore {v_v} = 0$

When the ball is at highest point, the horizontal velocity can be calculated by –

$

{v_h} = v\cos \theta \\

{v_h} = v\cos {45^ \circ } \\

{v_h} = \dfrac{v}{{\sqrt 2 }} \\

$

We know that, $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$

Now, from equation $\left( 1 \right)$

$E = \dfrac{1}{2}m{v^2}$

From the formula, we can see that kinetic energy is directly proportional to square of velocity. So,

$E\propto {v^2}$

If the velocity of the ball decreases by the factor of $\sqrt 2 $ then, kinetic energy will decrease by the factor of $2$.

Therefore, the kinetic energy at the highest point will be $\dfrac{E}{2}$.

**Hence, the correct option is (D).**

**Note**The kinetic energy of an object can be defined as the energy of an object which is possessed by the object during its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. The body can maintain the same kinetic energy until the speed of the object is changed. The S.I unit of kinetic energy is Joule which is denoted by $J$.

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