A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)

Options

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

Answer

To solve the problem step by step, we need to find the range of a projectile when its range is twice the greatest height attained. Let's denote the following:

- \( v \) = initial velocity of the projectile
- \( g \) = acceleration due to gravity
- \( R \) = range of the projectile
- \( H \) = greatest height attained by the projectile
- \( \theta \) = angle of projection

### Step 1: Write the formulas for range and height
The formulas for the range \( R \) and the maximum height \( H \) of a projectile are given by:

\[ R = \frac{v^2 \sin 2\theta}{g} \]

\[ H = \frac{v^2 \sin^2 \theta}{2g} \]

### Step 2: Set up the relationship between range and height
According to the problem, the range \( R \) is twice the greatest height \( H \):

\[ R = 2H \]

### Step 3: Substitute the formulas into the relationship
Substituting the formulas for \( R \) and \( H \) into the equation:

\[ \frac{v^2 \sin 2\theta}{g} = 2 \left( \frac{v^2 \sin^2 \theta}{2g} \right) \]

### Step 4: Simplify the equation
The \( g \) and \( v^2 \) terms can be canceled out (assuming \( g \neq 0 \) and \( v \neq 0 \)):

\[ \sin 2\theta = 2 \sin^2 \theta \]

### Step 5: Use the double angle identity
Using the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):

\[ 2 \sin \theta \cos \theta = 2 \sin^2 \theta \]

### Step 6: Simplify further
Dividing both sides by 2 (assuming \( \sin \theta \neq 0 \)):

\[ \sin \theta \cos \theta = \sin^2 \theta \]

### Step 7: Rearranging the equation
Rearranging gives:

\[ \cos \theta = \sin \theta \]

### Step 8: Solve for \( \theta \)
This implies:

\[ \tan \theta = 1 \quad \Rightarrow \quad \theta = 45^\circ \]

### Step 9: Substitute \( \theta \) back to find the range
Now substitute \( \theta = 45^\circ \) back into the range formula:

\[ R = \frac{v^2 \sin 90^\circ}{g} = \frac{v^2}{g} \]

### Step 10: Find the final expression for the range
Since \( \sin 90^\circ = 1 \):

\[ R = \frac{v^2}{g} \]

### Conclusion
Thus, the range of the projectile is:

\[ R = \frac{v^2}{g} \]