For an object thrown at `45^(@)` to the horizontal, the maximum height H and horizontal range R are related as

Options

A. `R = 16H`
B. `R = 8H`
C. `R = 4H`
D. `R = 2H`

Explanation

To find the relationship between the maximum height \( H \) and the horizontal range \( R \) for an object thrown at an angle of \( 45^\circ \) to the horizontal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the formulas for maximum height and range**:
- The maximum height \( H \) for a projectile is given by the formula:
\[ H = \frac{u_0^2 \sin^2 \theta}{2g} \]
- The horizontal range \( R \) is given by the formula:
\[ R = \frac{u_0^2 \sin 2\theta}{g} \]

2. **Substitute \( \theta = 45^\circ \)**:
- For \( \theta = 45^\circ \):
- \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \sin 2 \times 45^\circ = \sin 90^\circ = 1 \)

3. **Calculate maximum height \( H \)**:
- Substitute \( \theta = 45^\circ \) into the height formula:
\[ H = \frac{u_0^2 \sin^2 45^\circ}{2g} = \frac{u_0^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u_0^2 \cdot \frac{1}{2}}{2g} = \frac{u_0^2}{4g} \]

4. **Calculate horizontal range \( R \)**:
- Substitute \( \theta = 45^\circ \) into the range formula:
\[ R = \frac{u_0^2 \sin 90^\circ}{g} = \frac{u_0^2 \cdot 1}{g} = \frac{u_0^2}{g} \]

5. **Relate \( R \) and \( H \)**:
- Now we have:
\[ H = \frac{u_0^2}{4g} \quad \text{and} \quad R = \frac{u_0^2}{g} \]
- To find the relationship between \( R \) and \( H \), we can take the ratio:
\[ \frac{R}{H} = \frac{\frac{u_0^2}{g}}{\frac{u_0^2}{4g}} = \frac{u_0^2}{g} \cdot \frac{4g}{u_0^2} = 4 \]
- Therefore, we can conclude:
\[ R = 4H \]

### Final Result:
The relationship between the horizontal range \( R \) and the maximum height \( H \) for an object thrown at an angle of \( 45^\circ \) is:
\[ R = 4H \]