If K is the kinetic energy of a projectile fired at an angle 45°, then what is the kinetic energy at the highest point.

Options

A. `(K)/(4)`
B. `(K)/(2)`
C. K
D. 2K

Explanation

To find the kinetic energy of a projectile at its highest point when it is fired at an angle of 45°, we can follow these steps:

### Step 1: Understand the Initial Kinetic Energy
The initial kinetic energy (K) of the projectile when it is fired can be expressed as:
\[ K = \frac{1}{2} m u^2 \]
where \( m \) is the mass of the projectile and \( u \) is the initial velocity.

### Step 2: Break Down the Initial Velocity
When the projectile is fired at an angle of 45°, the initial velocity \( u \) can be broken down into two components:
- Horizontal component (\( u_x \)):
\[ u_x = u \cos(45^\circ) = u \cdot \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}} \]
- Vertical component (\( u_y \)):
\[ u_y = u \sin(45^\circ) = u \cdot \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}} \]

### Step 3: Analyze the Motion at the Highest Point
At the highest point of the projectile's trajectory, the vertical component of the velocity (\( u_y \)) becomes zero because the projectile has stopped rising and is about to fall back down. However, the horizontal component (\( u_x \)) remains unchanged:
\[ u_x = \frac{u}{\sqrt{2}} \]

### Step 4: Calculate the Kinetic Energy at the Highest Point
The kinetic energy at the highest point can be calculated using the horizontal component of the velocity:
\[ K_{highest} = \frac{1}{2} m u_x^2 \]
Substituting \( u_x \):
\[ K_{highest} = \frac{1}{2} m \left(\frac{u}{\sqrt{2}}\right)^2 = \frac{1}{2} m \cdot \frac{u^2}{2} = \frac{1}{4} m u^2 \]

### Step 5: Relate it to the Initial Kinetic Energy
Since the initial kinetic energy \( K = \frac{1}{2} m u^2 \), we can express \( K_{highest} \) in terms of \( K \):
\[ K_{highest} = \frac{1}{4} m u^2 = \frac{1}{2} \left(\frac{1}{2} m u^2\right) = \frac{K}{2} \]

### Final Answer
Thus, the kinetic energy at the highest point is:
\[ K_{highest} = \frac{K}{2} \]

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