A ball is projected to attain the maximum range. If the height attained is H, the range is

Options

A. H
B. 2H
C. 4H
D. H/2

Explanation

To solve the problem of finding the range of a ball projected to attain the maximum range when a certain height \( H \) is attained, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Maximum Range**:
The maximum range \( R_{\text{max}} \) for projectile motion occurs when the angle of projection \( \theta \) is \( 45^\circ \). The formula for maximum range is given by:
\[ R_{\text{max}} = \frac{u^2 \sin(2\theta)}{g} \]
where \( u \) is the initial velocity and \( g \) is the acceleration due to gravity.

2. **Calculating \( R_{\text{max}} \)**:
Since \( \theta = 45^\circ \), we have:
\[ \sin(2\theta) = \sin(90^\circ) = 1 \]
Therefore, the equation simplifies to:
\[ R_{\text{max}} = \frac{u^2}{g} \]

3. **Finding Maximum Height \( H \)**:
The maximum height \( H \) attained by the projectile can be calculated using the formula:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]
For \( \theta = 45^\circ \):
\[ \sin(45^\circ) = \frac{1}{\sqrt{2}} \quad \text{and} \quad \sin^2(45^\circ) = \frac{1}{2} \]
Substituting this into the height formula gives:
\[ H = \frac{u^2 \left(\frac{1}{2}\right)}{2g} = \frac{u^2}{4g} \]

4. **Relating Range and Height**:
From the expression for height \( H \):
\[ H = \frac{u^2}{4g} \]
We can express \( u^2 \) in terms of \( H \):
\[ u^2 = 4gH \]

5. **Substituting Back to Find Range**:
Now substituting \( u^2 \) back into the range formula:
\[ R_{\text{max}} = \frac{u^2}{g} = \frac{4gH}{g} = 4H \]

6. **Conclusion**:
Therefore, the range \( R \) when the height \( H \) is attained is given by:
\[ R = 4H \]

### Final Answer:
The range \( R \) is \( 4H \).

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