A particle with a velcoity (u) so that its horizontal ange is twice the greatest height attained. Find the horizontal range of it.

Options

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

Explanation

To solve the problem of finding the horizontal range of a particle whose horizontal angle is twice the greatest height attained, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between range and height**:
We are given that the horizontal range \( R \) is twice the greatest height \( H \) attained by the particle:
\[ R = 2H \]

2. **Write the formulas for range and height**:
The formula for the horizontal 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:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]

3. **Substitute the height formula into the range equation**:
Since \( R = 2H \), we can substitute \( H \) into this equation:
\[ R = 2 \left( \frac{u^2 \sin^2 \theta}{2g} \right) \]
Simplifying this gives:
\[ R = \frac{u^2 \sin^2 \theta}{g} \]

4. **Equate the two expressions for range**:
Now we have two expressions for \( R \):
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{g} \]
We can cancel \( \frac{u^2}{g} \) from both sides (assuming \( u \neq 0 \)):
\[ \sin 2\theta = \sin^2 \theta \]

5. **Use the identity for sine**:
We know that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Thus, we can rewrite the equation:
\[ 2 \sin \theta \cos \theta = \sin^2 \theta \]

6. **Rearranging the equation**:
Rearranging gives:
\[ 2 \cos \theta = \sin \theta \]
Dividing both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ 2 = \tan \theta \]

7. **Finding sine and cosine values**:
From \( \tan \theta = 2 \), we can represent this in a right triangle where the opposite side is 2 and the adjacent side is 1. The hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{2^2 + 1^2} = \sqrt{5} \]
Thus, we find:
\[ \sin \theta = \frac{2}{\sqrt{5}}, \quad \cos \theta = \frac{1}{\sqrt{5}} \]

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

9. **Final result**:
Therefore, the horizontal range \( R \) is:
\[ R = \frac{4u^2}{5g} \]

### Conclusion:
The final answer is:
\[ R = \frac{4u^2}{5g} \]