A projectile fired with initial velocity u at some angle `theta` has a range R . If the initial velocity be doubled at the same angle of projection, then the range will be

Options

A. `2R`
B. `R//2`
C. `R`
D. `4R`

Explanation

To solve the problem, we will analyze the relationship between the initial velocity of a projectile and its range.

### Step-by-Step Solution:

1. **Understanding the Range Formula**:
The range \( R \) of a projectile launched with an initial velocity \( u \) at an angle \( \theta \) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
where \( g \) is the acceleration due to gravity.

2. **Initial Conditions**:
In the initial scenario, the range is denoted as \( R \):
\[ R = \frac{u^2 \sin(2\theta)}{g} \]

3. **Doubling the Initial Velocity**:
Now, if the initial velocity is doubled, the new initial velocity \( u' \) becomes:
\[ u' = 2u \]
The angle of projection remains the same, \( \theta' = \theta \).

4. **Calculating the New Range**:
We need to find the new range \( R' \) with the new initial velocity:
\[ R' = \frac{(u')^2 \sin(2\theta')}{g} \]
Substituting \( u' = 2u \) and \( \theta' = \theta \):
\[ R' = \frac{(2u)^2 \sin(2\theta)}{g} \]
Simplifying this:
\[ R' = \frac{4u^2 \sin(2\theta)}{g} \]

5. **Relating New Range to Initial Range**:
Notice that the term \( \frac{u^2 \sin(2\theta)}{g} \) is exactly the original range \( R \):
\[ R' = 4 \cdot \frac{u^2 \sin(2\theta)}{g} = 4R \]

6. **Conclusion**:
Therefore, when the initial velocity is doubled, the new range \( R' \) becomes:
\[ R' = 4R \]

### Final Answer:
The new range when the initial velocity is doubled at the same angle of projection is \( 4R \).