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 maximum range given the height \( H \), we can follow these steps:

### Step 1: Understand the relationship between range and height
When a projectile is launched at an angle of \( 45^\circ \), it achieves the maximum range. The range \( R \) can be expressed in terms of the initial velocity \( u \) and the angle of projection \( \theta \) as follows:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
For \( \theta = 45^\circ \), \( \sin(90^\circ) = 1 \), thus:
\[ R_{\text{max}} = \frac{u^2}{g} \]

### Step 2: Calculate the maximum height
The maximum height \( H \) attained by the projectile can be expressed as:
\[ H = \frac{u^2 \sin^2(\theta)}{2g} \]
Again, substituting \( \theta = 45^\circ \) where \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ H = \frac{u^2 \left(\frac{1}{\sqrt{2}}\right)^2}{2g} = \frac{u^2 \cdot \frac{1}{2}}{2g} = \frac{u^2}{4g} \]

### Step 3: Relate range to height
From the expression for height, we can express \( u^2 \) in terms of \( H \):
\[ u^2 = 4gH \]

### Step 4: Substitute back into the range formula
Now, substituting \( u^2 \) back into the range formula:
\[ R_{\text{max}} = \frac{u^2}{g} = \frac{4gH}{g} = 4H \]

### Conclusion
Thus, the range \( R \) of the ball projected to attain maximum range when the height attained is \( H \) is:
\[ R = 4H \]