An object is projected with speed u and range of the projectile is found to be double of the maximum height attained by it . Range of the projectile is .

Options

A. `v_(A)=v_(B)=v_(C)`
B. `v_(A) gt v_(B) gt v_(C)`
C. `v_(A) lt v_(B) lt v_(C)`
D. `v_(a)=v_(B) gt v_(C)`

Explanation

To solve the problem, we need to derive the relationship between the range and the maximum height of a projectile. We are given that the range (R) is double the maximum height (H) attained by the projectile.

### Step-by-Step Solution:

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

2. **Set up the relationship between range and maximum height**:
- According to the problem, the range is double the maximum height:
\[ R = 2H \]

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

4. **Simplify the equation**:
- The \( g \) cancels out from both sides, and we can simplify:
\[ u^2 \sin 2\theta = u^2 \sin^2 \theta \]
- Dividing both sides by \( u^2 \) (assuming \( u \neq 0 \)):
\[ \sin 2\theta = \sin^2 \theta \]

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

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

7. **Find the solutions**:
- The solutions are:
- \( \sin \theta = 0 \) (not valid for projectile motion)
- \( \sin \theta - 2 \cos \theta = 0 \) or \( \sin \theta = 2 \cos \theta \)

8. **Using the tangent function**:
- Dividing by \( \cos \theta \):
\[ \tan \theta = 2 \]

9. **Find the range using the angle**:
- Using \( \tan \theta = 2 \), we can find \( \sin \theta \) and \( \cos \theta \):
\[ \sin \theta = \frac{2}{\sqrt{5}}, \quad \cos \theta = \frac{1}{\sqrt{5}} \]
- Substitute back into the range formula:
\[ R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{u^2 \cdot 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}}{g} = \frac{u^2 \cdot \frac{4}{5}}{g} \]
- Therefore, the range \( R \) is:
\[ R = \frac{4u^2}{5g} \]

### Final Answer:
The range of the projectile is:
\[ R = \frac{4u^2}{5g} \]