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'`

Answer

To solve the problem, we need to find the angle of projection (θ) given that the horizontal range (R) is three times the greatest height (H) of the projectile.

### Step-by-Step Solution:

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

2. **Setting Up the Relationship**:
- According to the problem, the range is three times the height:
\[ R = 3H \]

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

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

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

6. **Rearranging the Equation**:
- Rearranging gives:
\[ 4 \sin \theta \cos \theta = 3 \sin^2 \theta \]
- Dividing both sides by \(\sin \theta\) (assuming \(\sin \theta \neq 0\)):
\[ 4 \cos \theta = 3 \sin \theta \]

7. **Finding the Tangent**:
- Rearranging gives:
\[ \frac{\sin \theta}{\cos \theta} = \frac{4}{3} \]
- Therefore, we have:
\[ \tan \theta = \frac{4}{3} \]

8. **Calculating the Angle**:
- To find the angle θ, we take the arctangent:
\[ \theta = \tan^{-1}\left(\frac{4}{3}\right) \]
- Using a calculator, we find:
\[ \theta \approx 53.13^\circ \]

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