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

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `76^(@)`

Answer

To solve the problem, we need to find the angle of projection (θ) given that the velocity at the maximum height of a projectile is half of its initial velocity of projection (u).

### Step-by-Step Solution:

1. **Understand the Problem**: We know that at the maximum height of a projectile, the vertical component of the velocity becomes zero, and only the horizontal component remains. The horizontal component of the initial velocity (u) is given by \( u \cos(\theta) \).

2. **Set Up the Equation**: According to the problem, the velocity at the maximum height is half of the initial velocity. Therefore, we can write:
\[ u \cos(\theta) = \frac{u}{2} \]

3. **Simplify the Equation**: We can divide both sides of the equation by \( u \) (assuming \( u \neq 0 \)):
\[ \cos(\theta) = \frac{1}{2} \]

4. **Find the Angle**: Now, we need to find the angle \( \theta \) for which the cosine is \( \frac{1}{2} \). The angle whose cosine is \( \frac{1}{2} \) is:
\[ \theta = \cos^{-1}\left(\frac{1}{2}\right) \]

5. **Calculate the Value**: From trigonometric values, we know:
\[ \theta = 60^\circ \]

6. **Conclusion**: Therefore, the angle of projection is \( 60^\circ \).

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

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