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`

Answer

To solve the problem, we need to understand how the range of a projectile is affected by its initial velocity and the angle of projection.

### 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 Range Calculation**:
For the initial conditions, the range \( R \) can be expressed as:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
This is our initial range \( r \).

3. **Doubling the Initial Velocity**:
Now, if we double the initial velocity, the new initial velocity \( u' \) becomes:
\[ u' = 2u \]

4. **New Range Calculation**:
Using the same angle \( \theta \), the new range \( R' \) when the velocity is doubled is:
\[ R' = \frac{(u')^2 \sin(2\theta)}{g} = \frac{(2u)^2 \sin(2\theta)}{g} \]

5. **Simplifying the New Range**:
Simplifying the expression for \( R' \):
\[ R' = \frac{4u^2 \sin(2\theta)}{g} \]

6. **Relating New Range to Initial Range**:
We know from our initial range \( r \):
\[ r = \frac{u^2 \sin(2\theta)}{g} \]
Therefore, we can express the new range \( R' \) in terms of the initial range \( r \):
\[ R' = 4 \cdot \frac{u^2 \sin(2\theta)}{g} = 4r \]

7. **Final Conclusion**:
Thus, when the initial velocity is doubled while keeping the angle of projection the same, the new range becomes:
\[ R' = 4r \]

### Final Answer:
The new range will be \( 4R \).