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 a velocity \( v \) such that its range is twice the greatest height attained. Let's denote the range as \( R \) and the maximum height as \( H \).

### Step-by-Step Solution:

1. **Understand the relationship between range and height**:
We know that the range \( R \) is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
And the maximum height \( H \) is given by:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
According to the problem, we have:
\[ R = 2H \]

2. **Substituting the formulas**:
Substitute the expressions for \( R \) and \( H \) into the equation \( R = 2H \):
\[ \frac{u^2 \sin 2\theta}{g} = 2 \left( \frac{u^2 \sin^2 \theta}{2g} \right) \]
Simplifying this gives:
\[ \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{g} \]

3. **Cancelling common terms**:
Since \( u^2 \) and \( g \) are common on both sides, we can cancel them out (assuming \( u \neq 0 \) and \( g \neq 0 \)):
\[ \sin 2\theta = \sin^2 \theta \]

4. **Using the double angle identity**:
Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Thus, we can rewrite the equation as:
\[ 2 \sin \theta \cos \theta = \sin^2 \theta \]

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

6. **Finding solutions**:
This gives us two cases:
- Case 1: \( \sin \theta = 0 \) (which is not valid for projectile motion)
- Case 2: \( \sin \theta - 2 \cos \theta = 0 \) or \( \sin \theta = 2 \cos \theta \)

7. **Using the identity \( \tan \theta \)**:
From \( \sin \theta = 2 \cos \theta \), we can write:
\[ \tan \theta = 2 \]

8. **Finding \( \sin \theta \) and \( \cos \theta \)**:
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
Let \( \sin \theta = 2k \) and \( \cos \theta = k \):
\[ (2k)^2 + k^2 = 1 \implies 4k^2 + k^2 = 1 \implies 5k^2 = 1 \implies k = \frac{1}{\sqrt{5}} \]
Thus, \( \sin \theta = \frac{2}{\sqrt{5}} \) and \( \cos \theta = \frac{1}{\sqrt{5}} \).

9. **Calculating the range**:
Now substituting back into the range formula:
\[ R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \cdot 2 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}}}{g} = \frac{u^2 \cdot \frac{4}{5}}{g} \]
Therefore, the range \( R \) is:
\[ R = \frac{4u^2}{5g} \]

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