A projectile has a time of flight `T` and range `R`. If the time of flight is doubled, keeping the angle of projection same, what happens to the range ?

Options

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

Explanation

To solve the problem, we need to analyze the relationship between the time of flight \( T \) and the range \( R \) of a projectile.

### Step-by-Step Solution:

1. **Understand the formulas**:
- The time of flight \( T \) for a projectile launched at an angle \( \theta \) with initial velocity \( u \) is given by:
\[ T = \frac{2u \sin \theta}{g} \]
- The range \( R \) of the projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]

2. **Relate time of flight to range**:
- We can express \( u \) in terms of \( T \):
\[ u = \frac{gT}{2\sin \theta} \]
- Substitute this expression for \( u \) into the range formula:
\[ R = \frac{\left(\frac{gT}{2\sin \theta}\right)^2 \sin 2\theta}{g} \]
- Simplifying this gives:
\[ R = \frac{g^2 T^2}{4 \sin^2 \theta} \cdot \frac{2\sin \theta \cos \theta}{g} \]
\[ R = \frac{g T^2 \sin \theta \cos \theta}{2 \sin^2 \theta} \]
\[ R = \frac{g T^2}{2 \tan \theta} \]

3. **Determine the effect of doubling the time of flight**:
- If the time of flight \( T \) is doubled, we denote the new time of flight as \( T' = 2T \).
- Substitute \( T' \) into the range formula:
\[ R' = \frac{g (2T)^2}{2 \tan \theta} \]
\[ R' = \frac{g \cdot 4T^2}{2 \tan \theta} \]
\[ R' = 2 \cdot \frac{g T^2}{\tan \theta} = 4R \]

4. **Conclusion**:
- Therefore, if the time of flight is doubled while keeping the angle of projection the same, the range becomes \( 4R \).

### Final Answer:
The range becomes \( 4R \).