A particle is projected with a velocity `v` so that its range on a horizontal plane is twice the greatest height attained. If `g` is acceleration due to gravity, then its range is

Options

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

Explanation

To solve the problem, we need to find the range of a particle projected with an initial velocity \( v \) such that its range \( R \) is twice the greatest height \( H \) attained.

### Step-by-Step Solution:

1. **Identify the Relationship Between Range and Height**:
We are given that the range \( R \) is twice the greatest height \( H \):
\[ R = 2H \]

2. **Formulas for Range and Height**:
The formulas for the range \( R \) and the maximum height \( H \) for a projectile launched at an angle \( \theta \) with an initial velocity \( v \) are:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
\[ H = \frac{v^2 \sin^2 \theta}{2g} \]

3. **Substituting Height into the Range Equation**:
From the relationship \( R = 2H \), we can substitute for \( H \):
\[ R = 2 \left( \frac{v^2 \sin^2 \theta}{2g} \right) = \frac{v^2 \sin^2 \theta}{g} \]

4. **Equating the Two Expressions for Range**:
Now we have two expressions for \( R \):
\[ \frac{v^2 \sin 2\theta}{g} = \frac{v^2 \sin^2 \theta}{g} \]

5. **Simplifying the Equation**:
We can cancel \( \frac{v^2}{g} \) from both sides (assuming \( v \neq 0 \)):
\[ \sin 2\theta = \sin^2 \theta \]

6. **Using the Identity for Sine**:
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ 2 \sin \theta \cos \theta = \sin^2 \theta \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ \sin^2 \theta - 2 \sin \theta \cos \theta = 0 \]
Factoring out \( \sin \theta \):
\[ \sin \theta (\sin \theta - 2 \cos \theta) = 0 \]

8. **Finding Possible Angles**:
This gives us two cases:
- \( \sin \theta = 0 \) (which is not valid for projectile motion)
- \( \sin \theta = 2 \cos \theta \) or \( \tan \theta = 2 \)

9. **Finding the Range**:
We can now express \( \sin \theta \) and \( \cos \theta \) in terms of \( \tan \theta \):
Let \( \tan \theta = 2 \), then:
\[ \sin \theta = \frac{2}{\sqrt{5}}, \quad \cos \theta = \frac{1}{\sqrt{5}} \]

10. **Substituting Back to Find Range**:
Substitute \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ \sin 2\theta = 2 \left(\frac{2}{\sqrt{5}}\right) \left(\frac{1}{\sqrt{5}}\right) = \frac{4}{5} \]
Now substitute into the range formula:
\[ R = \frac{v^2 \cdot \frac{4}{5}}{g} = \frac{4v^2}{5g} \]

### Final Answer:
Thus, the range \( R \) is:
\[ R = \frac{4v^2}{5g} \]