Two bodies are projected from the same point with same velocity but in different directions. If the range in each case be Rand the times of flight be t and t' then R=... in terms of t,t'

Answer

To solve the problem, we need to derive the relationship between the range \( R \) of two projectiles launched at angles \( \theta \) and \( 90^\circ - \theta \) with the same initial velocity \( u \), in terms of their times of flight \( t \) and \( t' \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Two bodies are projected from the same point with the same initial velocity \( u \) but at different angles \( \theta \) and \( 90^\circ - \theta \).
- The range for both projectiles is the same, denoted as \( R \).
- The times of flight for the two projectiles are \( t \) and \( t' \).

2. **Formula for Range**:
- The formula for the range \( R \) of a projectile is given by:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- For the second projectile (launched at \( 90^\circ - \theta \)):
\[ R = \frac{u^2 \sin(2(90^\circ - \theta))}{g} = \frac{u^2 \sin(180^\circ - 2\theta)}{g} = \frac{u^2 \sin 2\theta}{g} \]
- Since both ranges are equal, we can use the same expression for both.

3. **Formula for Time of Flight**:
- The time of flight \( t \) for the first projectile is given by:
\[ t = \frac{2u \sin \theta}{g} \]
- The time of flight \( t' \) for the second projectile is:
\[ t' = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos \theta}{g} \]

4. **Relating Time of Flight to Range**:
- From the time of flight formulas, we can express \( u \sin \theta \) and \( u \cos \theta \) in terms of \( t \) and \( t' \):
\[ u \sin \theta = \frac{gt}{2} \]
\[ u \cos \theta = \frac{gt'}{2} \]

5. **Substituting into the Range Formula**:
- Now, substituting \( u \sin \theta \) and \( u \cos \theta \) into the range formula:
\[ R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 (2 \sin \theta \cos \theta)}{g} \]
- Replacing \( u \sin \theta \) and \( u \cos \theta \):
\[ R = \frac{(u \sin \theta)(u \cos \theta) \cdot 2}{g} = \frac{\left(\frac{gt}{2}\right)\left(\frac{gt'}{2}\right) \cdot 2}{g} \]
- Simplifying this gives:
\[ R = \frac{gtt'}{4} \]

6. **Final Expression**:
- Thus, the relationship between the range \( R \) and the times of flight \( t \) and \( t' \) is:
\[ R = \frac{g}{4} tt' \]

### Final Answer:
\[ R = \frac{g}{4} tt' \]