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 )/(2)`
C. `(E )/(sqrt2)`
D. 0

Answer

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

### Step-by-Step Solution:

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

2. **Components of Initial Velocity:**
When the ball is projected at an angle of 45 degrees, we can resolve its initial velocity \( U \) into horizontal and vertical components:
- Horizontal component \( U_x = U \cos(45^\circ) = U \cdot \frac{1}{\sqrt{2}} \)
- Vertical component \( U_y = U \sin(45^\circ) = U \cdot \frac{1}{\sqrt{2}} \)

3. **Velocity at the Highest Point:**
At the highest point of its trajectory, the vertical component of the velocity becomes zero (\( V_y = 0 \)). Therefore, the only component of velocity at this point is the horizontal component:
\[ V_x = U \cos(45^\circ) = U \cdot \frac{1}{\sqrt{2}} \]

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

5. **Relating Final Kinetic Energy to Initial Kinetic Energy:**
From the initial kinetic energy \( E = \frac{1}{2} m U^2 \), we can express \( m U^2 \) as:
\[ m U^2 = 2E \]
Substituting this into the expression for \( KE_{final} \):
\[ KE_{final} = \frac{1}{4} (2E) = \frac{E}{2} \]

6. **Final Answer:**
Thus, the kinetic energy of the ball at the highest point of its flight is:
\[ \boxed{\frac{E}{2}} \]