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

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. None

Answer

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

### Step-by-Step Solution:

1. **Understand the formulas:**
- The formula for the range (R) of a projectile is:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- The formula for the maximum height (H) of a projectile is:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Set up the equation based on the given condition:**
- According to the problem, the range is \(4\sqrt{3}\) times the maximum height:
\[ R = 4\sqrt{3} H \]

3. **Substituting the formulas into the equation:**
- 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. **Simplify the equation:**
- Cancel \(g\) and \(u^2\) from both sides (assuming \(g \neq 0\) and \(u \neq 0\)):
\[ \sin 2\theta = 4\sqrt{3} \cdot \frac{1}{2} \sin^2 \theta \]
- This simplifies to:
\[ \sin 2\theta = 2\sqrt{3} \sin^2 \theta \]

5. **Using the double angle identity:**
- Recall that \(\sin 2\theta = 2 \sin \theta \cos \theta\):
\[ 2 \sin \theta \cos \theta = 2\sqrt{3} \sin^2 \theta \]

6. **Dividing both sides by 2:**
- We can simplify further:
\[ \sin \theta \cos \theta = \sqrt{3} \sin^2 \theta \]

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

8. **Dividing both sides by \(\sin \theta\) (assuming \(\sin \theta \neq 0\)):**
- This leads to:
\[ \cot \theta = \sqrt{3} \]

9. **Finding the angle:**
- The angle whose cotangent is \(\sqrt{3}\) is:
\[ \theta = 30^\circ \]

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