A body is projected at such an angle that the horizontal range is three times the greatest height. The-angle of projection is

Options

A. `25^(@)8'`
B. `33^(@)7'`
C. `42^(@)8'`
D. `53^(@)8'`

Explanation

To solve the problem, we need to find the angle of projection (θ) at which the horizontal range (R) is three times the maximum height (H) of the projectile. We will use the formulas for range and maximum height in projectile motion.

### Step-by-Step Solution:

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

2. **Setting Up the Relationship**:
- According to the problem, we know that:
\[ R = 3H \]
- Substituting the formulas for \( R \) and \( H \) into this equation gives:
\[ \frac{u^2 \sin(2\theta)}{g} = 3 \left( \frac{u^2 \sin^2(\theta)}{2g} \right) \]

3. **Simplifying the Equation**:
- We can cancel \( u^2 \) and \( g \) from both sides (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \sin(2\theta) = \frac{3}{2} \sin^2(\theta) \]

4. **Using the Double Angle Identity**:
- Recall that \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ 2 \sin(\theta) \cos(\theta) = \frac{3}{2} \sin^2(\theta) \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ 4 \cos(\theta) = 3 \sin(\theta) \]

6. **Dividing by \( \cos(\theta) \)**:
- Dividing both sides by \( \cos(\theta) \) (assuming \( \cos(\theta) \neq 0 \)):
\[ 4 = 3 \tan(\theta) \]

7. **Finding the Angle**:
- Solving for \( \tan(\theta) \):
\[ \tan(\theta) = \frac{4}{3} \]
- Therefore, the angle \( \theta \) can be found using the arctangent function:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]

8. **Calculating the Angle**:
- Using a calculator, we find:
\[ \theta \approx 53.13^\circ \]
- This can be converted to degrees and minutes:
\[ \theta \approx 53^\circ 8' \]

### Final Answer:
The angle of projection is approximately \( 53^\circ 8' \).