An object projected with same speed at two different angles covers the same horizontal range R. If the two times of flight be ` t_(1) and t_(2)` . The range is ` 1/alpha "gt"_(1) t_(2), ` the value of `alpha` is

Answer

To solve the problem, we need to analyze the projectile motion of an object projected at two different angles but with the same initial speed, resulting in the same horizontal range \( R \). We will derive the relationship between the time of flight \( t_1 \) and \( t_2 \) and find the value of \( \alpha \).

### Step-by-Step Solution:

1. **Understanding the Angles**:
- Let the two angles of projection be \( \theta_1 \) and \( \theta_2 \).
- Since the ranges are the same, \( \theta_1 \) and \( \theta_2 \) are complementary angles, which means:
\[ \theta_1 + \theta_2 = 90^\circ \]

2. **Expression for Range**:
- The range \( R \) for a projectile is given by the formula:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
- For our two angles, we have:
\[ R = \frac{u^2 \sin 2\theta_1}{g} = \frac{u^2 \sin 2\theta_2}{g} \]

3. **Time of Flight**:
- The time of flight for the projectile at angle \( \theta_1 \) is:
\[ t_1 = \frac{2u \sin \theta_1}{g} \]
- The time of flight for the projectile at angle \( \theta_2 \) is:
\[ t_2 = \frac{2u \sin \theta_2}{g} \]

4. **Using Complementary Angles**:
- Since \( \theta_2 = 90^\circ - \theta_1 \):
\[ \sin \theta_2 = \cos \theta_1 \]
- Therefore:
\[ t_2 = \frac{2u \cos \theta_1}{g} \]

5. **Relating Range to Time of Flight**:
- We can express \( \sin \theta_1 \) and \( \cos \theta_1 \) in terms of \( t_1 \) and \( t_2 \):
\[ \sin \theta_1 = \frac{gt_1}{2u} \]
\[ \cos \theta_1 = \frac{gt_2}{2u} \]

6. **Substituting into Range Formula**:
- The range can also be expressed using the sine and cosine relationships:
\[ R = \frac{u^2 \sin 2\theta_1}{g} = \frac{u^2 (2 \sin \theta_1 \cos \theta_1)}{g} \]
- Substituting for \( \sin \theta_1 \) and \( \cos \theta_1 \):
\[ R = \frac{u^2 \cdot 2 \left(\frac{gt_1}{2u}\right) \left(\frac{gt_2}{2u}\right)}{g} \]
- Simplifying this gives:
\[ R = \frac{g t_1 t_2}{2u} \]

7. **Comparing with Given Expression**:
- The problem states that:
\[ R = \frac{1}{\alpha} g t_1 t_2 \]
- From our derived expression, we have:
\[ R = \frac{g t_1 t_2}{2u} \]
- Equating the two expressions:
\[ \frac{g t_1 t_2}{2u} = \frac{1}{\alpha} g t_1 t_2 \]
- Cancelling \( g t_1 t_2 \) from both sides (assuming \( t_1 \) and \( t_2 \) are not zero):
\[ \frac{1}{2u} = \frac{1}{\alpha} \]
- Therefore:
\[ \alpha = 2 \]

### Final Answer:
The value of \( \alpha \) is \( 2 \).