A ball whose kinetic energy is `E` , is projected at an angle of `45^@` to the horizontal . The kinetic energy of the ball at the highest point of its flight will be

Options

A. `E`
B. `(E)/sqrt(2)`
C. `(E)/(2)`
D. `zero`

Explanation

To solve the problem, we need to analyze the kinetic energy of a ball projected at an angle of \(45^\circ\) to the horizontal at the highest point of its flight.

### Step-by-Step Solution:

1. **Understanding Initial Kinetic Energy**:
The initial kinetic energy \(E\) of the ball when it is projected can be expressed as:
\[ E = \frac{1}{2} m U^2 \]
where \(U\) is the initial velocity of the ball and \(m\) is its mass.

2. **Components of Initial Velocity**:
When the ball is projected at an angle of \(45^\circ\), we can break the initial velocity \(U\) into its horizontal and vertical components:
\[ U_x = U \cos(45^\circ) = \frac{U}{\sqrt{2}} \]
\[ U_y = U \sin(45^\circ) = \frac{U}{\sqrt{2}} \]

3. **Velocity at the Highest Point**:
At the highest point of its flight, the vertical component of the velocity becomes zero (as the ball stops rising at that point). Therefore, the only component of velocity remaining is the horizontal component:
\[ v = U_x = \frac{U}{\sqrt{2}} \]

4. **Calculating Kinetic Energy at the Highest Point**:
The kinetic energy at the highest point can be calculated using the horizontal component of the velocity:
\[ KE_{highest} = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{U}{\sqrt{2}}\right)^2 \]
Simplifying this expression:
\[ KE_{highest} = \frac{1}{2} m \left(\frac{U^2}{2}\right) = \frac{1}{4} m U^2 \]

5. **Relating to Initial Kinetic Energy**:
We know from the initial kinetic energy expression that:
\[ E = \frac{1}{2} m U^2 \]
Therefore, we can express the kinetic energy at the highest point in terms of \(E\):
\[ KE_{highest} = \frac{1}{4} m U^2 = \frac{1}{2} \left(\frac{1}{2} m U^2\right) = \frac{1}{2} E \]

### Final Answer:
The kinetic energy of the ball at the highest point of its flight will be:
\[ \frac{E}{2} \]