The horizontal range is four times the maximum height attained by a projectile. The angle of projection is

Options

A. `90^(@)`
B. `60^(@)`
C. `45^(@)`
D. `30^(@)`

Answer

To solve the problem, we need to find the angle of projection \(\theta\) given that the horizontal range \(R\) is four times the maximum height \(H\) attained by a projectile.

### Step-by-step Solution:

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

2. **Set up the relationship**:
- According to the problem, the horizontal range is four times the maximum height:
\[ R = 4H \]

3. **Substitute the formulas into the relationship**:
- Replace \(R\) and \(H\) in the equation:
\[ \frac{U^2 \sin 2\theta}{g} = 4 \left(\frac{U^2 \sin^2 \theta}{2g}\right) \]

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

5. **Use the double angle identity**:
- Recall that \(\sin 2\theta = 2 \sin \theta \cos \theta\):
\[ 2 \sin \theta \cos \theta = 2 \sin^2 \theta \]

6. **Divide both sides by 2**:
- We can simplify further:
\[ \sin \theta \cos \theta = \sin^2 \theta \]

7. **Rearrange the equation**:
- Rearranging gives us:
\[ \sin \theta \cos \theta - \sin^2 \theta = 0 \]
- Factor out \(\sin \theta\):
\[ \sin \theta (\cos \theta - \sin \theta) = 0 \]

8. **Solve for \(\theta\)**:
- This gives us two cases:
1. \(\sin \theta = 0\) (which is not a valid angle of projection)
2. \(\cos \theta - \sin \theta = 0\) or \(\cos \theta = \sin \theta\)

- The angle where \(\cos \theta = \sin \theta\) is:
\[ \theta = 45^\circ \]

### Conclusion:
The angle of projection \(\theta\) is \(45^\circ\).