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. `(2u^(2))/(5g)`
B. `(3u^(2))/(5g)`
C. `(4u^(2))/(5g)`
D. `(u^(2))/(g)`

Explanation

To solve the problem, we need to find the range of a projectile given that the range \( R \) is double the maximum height \( H_{max} \) attained by it.

### Step-by-Step Solution:

1. **Understanding the Relationships**:
- The range \( R \) of a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- The maximum height \( H_{max} \) is given by:
\[ H_{max} = \frac{u^2 \sin^2 \theta}{2g} \]

2. **Setting Up the Equation**:
- According to the problem, the range is double the maximum height:
\[ R = 2H_{max} \]
- Substituting the formulas for \( R \) and \( H_{max} \):
\[ \frac{u^2 \sin 2\theta}{g} = 2 \left(\frac{u^2 \sin^2 \theta}{2g}\right) \]

3. **Simplifying the Equation**:
- The \( g \) cancels out from both sides:
\[ u^2 \sin 2\theta = u^2 \sin^2 \theta \]
- Dividing both sides by \( u^2 \) (assuming \( u \neq 0 \)):
\[ \sin 2\theta = \sin^2 \theta \]

4. **Using the Double Angle Identity**:
- Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ 2 \sin \theta \cos \theta = \sin^2 \theta \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ \sin^2 \theta - 2 \sin \theta \cos \theta = 0 \]
- Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 2 \cos \theta) = 0 \]
- This gives us two cases:
1. \( \sin \theta = 0 \) (not valid for projectile motion)
2. \( \sin \theta = 2 \cos \theta \)

6. **Finding \( \tan \theta \)**:
- From \( \sin \theta = 2 \cos \theta \), we can write:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = 2 \]

7. **Finding \( \sin \theta \) and \( \cos \theta \)**:
- Using \( \tan \theta = 2 \):
- Let \( \sin \theta = 2k \) and \( \cos \theta = k \) for some \( k \).
- Then \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ (2k)^2 + k^2 = 1 \implies 4k^2 + k^2 = 1 \implies 5k^2 = 1 \implies k^2 = \frac{1}{5} \implies k = \frac{1}{\sqrt{5}} \]
- Thus, \( \sin \theta = \frac{2}{\sqrt{5}} \) and \( \cos \theta = \frac{1}{\sqrt{5}} \).

8. **Substituting Back to Find the Range**:
- Now substituting \( \sin \theta \) and \( \cos \theta \) back into the range formula:
\[ R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 (2 \sin \theta \cos \theta)}{g} = \frac{u^2 \cdot 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}}{g} \]
\[ R = \frac{u^2 \cdot \frac{4}{5}}{g} = \frac{4u^2}{5g} \]

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