The horizontal range of a projectile is `4 sqrt(3)` times its maximum height. Its angle of projection will be

Options

A. `60^(@)`
B. `37^(@)`
C. `30^(@)`
D. `45^(@)`

Explanation

To solve the problem, we need to find the angle of projection (θ) of a projectile given that its horizontal range (R) is \(4\sqrt{3}\) times its maximum height (H).

### Step-by-Step Solution:

1. **Understand the Relationship**:
We know that the horizontal range \(R\) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
where \(u\) is the initial velocity, \(\theta\) is the angle of projection, and \(g\) is the acceleration due to gravity.

The maximum height \(H\) of a projectile is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Set Up the Equation**:
According to the problem, we have:
\[ R = 4\sqrt{3} H \]
Substituting the formulas for \(R\) and \(H\) into this equation gives:
\[ \frac{u^2 \sin 2\theta}{g} = 4\sqrt{3} \left(\frac{u^2 \sin^2 \theta}{2g}\right) \]

3. **Simplify the Equation**:
Cancel \(u^2\) and \(g\) from both sides:
\[ \sin 2\theta = 4\sqrt{3} \cdot \frac{\sin^2 \theta}{2} \]
This simplifies to:
\[ \sin 2\theta = 2\sqrt{3} \sin^2 \theta \]

4. **Use the Identity for \(\sin 2\theta\)**:
Recall that \(\sin 2\theta = 2 \sin \theta \cos \theta\). Substitute this into the equation:
\[ 2 \sin \theta \cos \theta = 2\sqrt{3} \sin^2 \theta \]

5. **Cancel Common Terms**:
We can divide both sides by 2 (assuming \(\sin \theta \neq 0\)):
\[ \sin \theta \cos \theta = \sqrt{3} \sin^2 \theta \]

6. **Rearrange the Equation**:
Divide both sides by \(\sin \theta\) (assuming \(\sin \theta \neq 0\)):
\[ \cos \theta = \sqrt{3} \sin \theta \]

7. **Express in Terms of \(\tan \theta\)**:
Rearranging gives:
\[ \tan \theta = \frac{\cos \theta}{\sin \theta} = \frac{1}{\sqrt{3}} \]

8. **Find the Angle**:
The angle \(\theta\) for which \(\tan \theta = \frac{1}{\sqrt{3}}\) is:
\[ \theta = 30^\circ \]

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