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 find the angle of projection (θ) of a projectile given that its kinetic energy at the highest point is half of its initial kinetic energy.

### Step-by-Step Solution:

1. **Understanding the Initial Kinetic Energy (KE_initial)**:
The initial kinetic energy of the projectile when it is launched is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 \]
where \( m \) is the mass of the projectile and \( u \) is the initial velocity.

2. **Components of Initial Velocity**:
When the projectile is launched at an angle \( \theta \), the components of the initial velocity are:
- Horizontal component: \( u \cos \theta \)
- Vertical component: \( u \sin \theta \)

3. **Velocity at the Highest Point**:
At the highest point of its trajectory, the vertical component of the velocity becomes zero. Therefore, the velocity at the highest point is only the horizontal component:
\[ v_{\text{highest}} = u \cos \theta \]

4. **Kinetic Energy at the Highest Point (KE_final)**:
The kinetic energy at the highest point is given by:
\[ KE_{\text{final}} = \frac{1}{2} m (u \cos \theta)^2 = \frac{1}{2} m u^2 \cos^2 \theta \]

5. **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, we have:
\[ \frac{1}{2} m u^2 \cos^2 \theta = \frac{1}{2} \left( \frac{1}{2} m u^2 \right) \]

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

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

8. **Finding the Angle θ**:
The angle \( \theta \) for which \( \cos \theta = \frac{1}{\sqrt{2}} \) is:
\[ \theta = 45^\circ \]

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

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