The kinetic energy of a projectile at the highest point is half of the initial kinetic energy. The angle of projection with the horizontal is

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `90^@`

Answer

To solve the problem, we need to determine the angle of projection (θ) for a projectile such that its kinetic energy at the highest point is half of its initial kinetic energy.

### Step-by-Step Solution:

1. **Understanding the Components of Velocity**:
- When a projectile is launched with an initial velocity \( v \) at an angle \( \theta \) with the horizontal, it has two components:
- Horizontal component: \( v_x = v \cos \theta \)
- Vertical component: \( v_y = v \sin \theta \)

2. **Kinetic Energy at the Initial Point**:
- The initial kinetic energy (KE_initial) of the projectile can be expressed as:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 \]
- Here, \( m \) is the mass of the projectile.

3. **Kinetic Energy at the Highest Point**:
- At the highest point of its trajectory, the vertical component of the velocity becomes zero (\( v_y = 0 \)), and only the horizontal component remains:
\[ v_x = v \cos \theta \]
- Therefore, the kinetic energy at the highest point (KE_final) is:
\[ KE_{\text{final}} = \frac{1}{2} m (v \cos \theta)^2 = \frac{1}{2} m v^2 \cos^2 \theta \]

4. **Setting Up the Equation**:
- According to the problem, the kinetic energy at the highest point is half of the initial kinetic energy:
\[ KE_{\text{final}} = \frac{1}{2} KE_{\text{initial}} \]
- Substituting the expressions for kinetic energy:
\[ \frac{1}{2} m v^2 \cos^2 \theta = \frac{1}{2} \left( \frac{1}{2} m v^2 \right) \]

5. **Simplifying the Equation**:
- Canceling out \( \frac{1}{2} m v^2 \) from both sides gives:
\[ \cos^2 \theta = \frac{1}{2} \]

6. **Finding the Value of Cosine**:
- Taking the square root of both sides:
\[ \cos \theta = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

7. **Determining the Angle**:
- The angle \( \theta \) that corresponds to \( \cos \theta = \frac{1}{\sqrt{2}} \) is:
\[ \theta = 45^\circ \]

### Final Answer:
The angle of projection with the horizontal is \( 45^\circ \).

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