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

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 Formulas**:
- The horizontal range (R) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- The maximum height (H) of a projectile is given by the formula:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Set Up the Relationship**:
- According to the problem, the horizontal range is \(4\sqrt{3}\) times the maximum height:
\[ R = 4\sqrt{3} H \]

3. **Substitute the Formulas**:
- Substitute the expressions for R and H into the equation:
\[ \frac{u^2 \sin 2\theta}{g} = 4\sqrt{3} \left(\frac{u^2 \sin^2 \theta}{2g}\right) \]

4. **Cancel Common Terms**:
- Cancel \(u^2\) and \(g\) from both sides (assuming \(u \neq 0\) and \(g \neq 0\)):
\[ \sin 2\theta = 4\sqrt{3} \cdot \frac{\sin^2 \theta}{2} \]
- This simplifies to:
\[ \sin 2\theta = 2\sqrt{3} \sin^2 \theta \]

5. **Use the Double Angle Identity**:
- Recall that \(\sin 2\theta = 2 \sin \theta \cos \theta\):
\[ 2 \sin \theta \cos \theta = 2\sqrt{3} \sin^2 \theta \]
- Dividing both sides by 2 (assuming \(\sin \theta \neq 0\)):
\[ \sin \theta \cos \theta = \sqrt{3} \sin^2 \theta \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ \cos \theta = \sqrt{3} \sin \theta \]

7. **Using the Tangent Function**:
- Dividing both sides by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)):
\[ 1 = \sqrt{3} \tan \theta \]
- Thus, we have:
\[ \tan \theta = \frac{1}{\sqrt{3}} \]

8. **Finding the Angle**:
- The angle whose tangent is \(\frac{1}{\sqrt{3}}\) is:
\[ \theta = 30^\circ \]

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