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

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To solve the problem, we need to analyze the motion of a projectile at its maximum height. Here's a step-by-step solution:

### Step 1: Understand the Components of Initial Velocity
When a projectile is launched with an initial speed \( u \) at an angle \( \theta \) with the horizontal, it has two components of velocity:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Analyze the Speed at Maximum Height
At the maximum height of the projectile's trajectory, the vertical component of the velocity becomes zero (\( u_y = 0 \)). Therefore, the speed of the projectile at maximum height is entirely due to the horizontal component:
\[ v = u_x = u \cos \theta \]

### Step 3: Use the Given Condition
According to the problem, the speed at maximum height is half of the initial speed:
\[ v = \frac{u}{2} \]

### Step 4: Set Up the Equation
From the previous steps, we have:
\[ u \cos \theta = \frac{u}{2} \]

### Step 5: Simplify the Equation
We can cancel \( u \) from both sides (assuming \( u \neq 0 \)):
\[ \cos \theta = \frac{1}{2} \]

### Step 6: Find the Angle \( \theta \)
The angle \( \theta \) that satisfies \( \cos \theta = \frac{1}{2} \) is:
\[ \theta = 60^\circ \]

### Conclusion
Thus, the angle of projection is \( 60^\circ \).

---