The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is .

Answer

To solve the problem, we need to determine the angle of projection (θ) when the speed of a projectile at its maximum height is half of its initial speed (u).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We know that the speed of the projectile at its maximum height is half of its initial speed.
- Let the initial speed be \( u \). Therefore, at maximum height, the speed \( v = \frac{u}{2} \).

2. **Components of Initial Velocity**:
- The initial velocity can be resolved into two components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

3. **Velocity at Maximum Height**:
- At the maximum height, the vertical component of the velocity becomes zero. Thus, the only component of velocity at this point is the horizontal component:
- \( v = u_x = u \cos \theta \)

4. **Setting Up the Equation**:
- Since we know that the speed at maximum height is half of the initial speed, we can set up the equation:
\[ u \cos \theta = \frac{u}{2} \]

5. **Simplifying the Equation**:
- We can cancel \( u \) from both sides (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{1}{2} \]

6. **Finding the Angle**:
- The value of \( \theta \) for which \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

### Final Answer:
The angle of projection \( \theta \) is \( 60^\circ \).