$ ext{The horizontal range of a projectile is } 4\sqrt{3} ext{ times of its maximum height. What will be the value of its projection angle ?}$

Q: $\text{The horizontal range of a projectile is } 4\sqrt{3} \text{ times of its maximum height. What will be the value of its projection angle ?}$

$\text{The horizontal range of a projectile is } 4\sqrt{3} \text{ times of its maximum height. What will be the value of its projection angle ?}$

Question

$	ext{The horizontal range of a projectile is } 4\sqrt{3} 	ext{ times of its maximum height. What will be the value of its projection angle ?}$

Accepted Answer

To determine why Option 3 (30°) is the correct answer for the given problem, we need to analyze the relationship between the horizontal range and the maximum height of a projectile.

### Explanation:

1. **Projectile Motion Basics:**
   - The horizontal range $ R $ of a projectile is given by the formula:
     $$
     R = \frac{v^2 \sin(2\theta)}{g}
     $$
   - The maximum height $ H $ is given by:
     $$
     H = \frac{v^2 \sin^2(\theta)}{2g}
     $$
   where $ v $ is the initial velocity, $ \theta $ is the angle of projection, and $ g $ is the acceleration due to gravity.

2. **Given Condition:**
   - The horizontal range is $ 4\sqrt{3} $ times the maximum height:
     $$
     R = 4\sqrt{3} \times H
     $$

3. **Substitute the Formulas:**
   - Substitute the formulas for $ R $ and $ H $ into the given condition:
     $$
     \frac{v^2 \sin(2\theta)}{g} = 4\sqrt{3} \times \frac{v^2 \sin^2(\theta)}{2g}
     $$
   - Simplify by canceling out common terms:
     $$
     \sin(2\theta) = 2\sqrt{3} \sin^2(\theta)
     $$

4. **Trigonometric Identity:**
   - Use the identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$:
     $$
     2\sin(\theta)\cos(\theta) = 2\sqrt{3} \sin^2(\theta)
     $$
   - Divide both sides by $ 2\sin(\theta) $ (assuming $\sin(\theta) \neq 0$):
     $$
     \cos(\theta) = \sqrt{3} \sin(\theta)
     $$

5. **Solve for $\theta$:**
   - Divide both sides by $\cos(\theta)$:
     $$
     1 = \sqrt{3} \tan(\theta)
     $$
   - Solve for $\tan(\theta)$:
     $$
     \tan(\theta) = \frac{1}{\sqrt{3}}
     $$
   - The angle $\theta$ that satisfies $\tan(\theta) = \frac{1}{\sqrt{3}}$ is 30°.

### Why Other Options are Incorrect:

1. **Option 1 (45°):**
   - At 45°, the range is maximized, but the condition $ R = 4\sqrt{3} \times H $ does not hold. For 45°, the range is 4 times the height, not $ 4\sqrt{3} $.

2. **Option 2 (60°):**
   - At 60°, the relationship between range and height does not satisfy the given condition. The calculations would not yield the factor of $ 4\sqrt{3} $.

3. **Option 4 (90°):**
   - At 90°, the projectile would go straight up and come straight down, resulting in zero horizontal range, which does not satisfy any condition involving a non-zero range.

Thus, the correct answer is indeed Option 3 (30°), as it satisfies the given condition that the horizontal range is $ 4\sqrt{3} $ times the maximum height.