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 when the angle of projection remains constant.

### Step-by-Step Solution:

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

2. **Square the time of flight formula**:
- To relate \( T \) and \( R \), we can square the time of flight:
\[ T^2 = \left(\frac{2u \sin \theta}{g}\right)^2 = \frac{4u^2 \sin^2 \theta}{g^2} \]
- This is our **Equation 1**.

3. **Express range in terms of time of flight**:
- We can express \( R \) in terms of \( T^2 \). From the range formula:
\[ R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 (2 \sin \theta \cos \theta)}{g} \]
- This is our **Equation 2**.

4. **Relate \( R \) and \( T^2 \)**:
- We can divide \( R \) by \( T^2 \):
\[ \frac{R}{T^2} = \frac{\frac{u^2 (2 \sin \theta \cos \theta)}{g}}{\frac{4u^2 \sin^2 \theta}{g^2}} \]
- Simplifying this gives:
\[ \frac{R}{T^2} = \frac{2g \cos \theta}{4 \sin \theta} = \frac{g \cos \theta}{2 \sin \theta} \]

5. **Determine the new range when time of flight is doubled**:
- If the time of flight is doubled, \( T \) becomes \( 2T \). Therefore, \( T^2 \) becomes \( (2T)^2 = 4T^2 \).
- Using the relationship we derived:
\[ R_2 = \frac{R}{T^2} \cdot (4T^2) = 4R \]
- Thus, the new range \( R_2 \) is four times the original range \( R \).

### Final Answer:
When the time of flight is doubled while keeping the angle of projection the same, the range becomes:
\[ R_2 = 4R \]