The velocity at the maximum height of a projectile is `sqrt(3)/2` times the initial velocity of projection. The angle of projection is

Options

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

Answer

To solve the problem, we need to find the angle of projection (\(\theta\)) given that the velocity at the maximum height of a projectile is \(\frac{\sqrt{3}}{2}\) times the initial velocity of projection (\(u\)).

### Step-by-Step Solution:

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

2. **Velocity at Maximum Height**:
- At the maximum height, the vertical component of the velocity becomes zero (\(u_y = 0\)), while the horizontal component remains constant throughout the motion.
- Therefore, the velocity at maximum height (\(v\)) is equal to the horizontal component:
\[ v = u_x = u \cos \theta \]

3. **Setting Up the Equation**:
- According to the problem, the velocity at the maximum height is given as:
\[ v = \frac{\sqrt{3}}{2} u \]
- Substituting the expression for \(v\) from step 2, we have:
\[ u \cos \theta = \frac{\sqrt{3}}{2} u \]

4. **Cancelling \(u\)**:
- Since \(u\) is not zero, we can divide both sides of the equation by \(u\):
\[ \cos \theta = \frac{\sqrt{3}}{2} \]

5. **Finding the Angle \(\theta\)**:
- To find \(\theta\), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \]
- The angle whose cosine is \(\frac{\sqrt{3}}{2}\) is \(30^\circ\):
\[ \theta = 30^\circ \]

### Final Answer:
The angle of projection is \(30^\circ\).

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