Two particls are projected in air with speed u at angles `theta_(1) and theta_(2)` (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then which one of the following is correct? where `T_(1) and T_(2)` are the time of flight.

Options

A. `theta_(1) gt theta_(2)`
B. `theta_(1) = theta_(2)`
C. `T_(1) lt T_(2)`
D. `T_(1) = T_(2)`

Explanation

To solve the problem, we need to analyze the conditions given for the two particles projected at different angles and determine the relationship between their angles and times of flight based on the heights they reach.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Two particles are projected with the same initial speed \( u \) at angles \( \theta_1 \) and \( \theta_2 \) to the horizontal.
- It is given that the height reached by the first particle \( h_1 \) is greater than the height reached by the second particle \( h_2 \).

2. **Height Formula**:
- The maximum height \( h \) reached by a projectile is given by the formula:
\[ h = \frac{u^2 \sin^2 \theta}{2g} \]
- For the two particles, we can write:
\[ h_1 = \frac{u^2 \sin^2 \theta_1}{2g} \]
\[ h_2 = \frac{u^2 \sin^2 \theta_2}{2g} \]

3. **Setting Up the Inequality**:
- Since \( h_1 > h_2 \), we can write:
\[ \frac{u^2 \sin^2 \theta_1}{2g} > \frac{u^2 \sin^2 \theta_2}{2g} \]
- Simplifying this, we find:
\[ \sin^2 \theta_1 > \sin^2 \theta_2 \]

4. **Taking Square Roots**:
- Since both angles are acute, we can take the square root of both sides without changing the inequality:
\[ \sin \theta_1 > \sin \theta_2 \]

5. **Relating Angles**:
- The sine function is increasing in the interval \( (0, \frac{\pi}{2}) \) (or \( (0, 90^\circ) \)), therefore:
\[ \theta_1 > \theta_2 \]

6. **Time of Flight Formula**:
- The time of flight \( T \) for a projectile is given by:
\[ T = \frac{2u \sin \theta}{g} \]
- For the two particles, we have:
\[ T_1 = \frac{2u \sin \theta_1}{g} \]
\[ T_2 = \frac{2u \sin \theta_2}{g} \]

7. **Comparing Times of Flight**:
- Since we established that \( \theta_1 > \theta_2 \) and knowing that sine is an increasing function, we can conclude:
\[ \sin \theta_1 > \sin \theta_2 \]
- Therefore:
\[ T_1 > T_2 \]

8. **Conclusion**:
- The correct relationship is \( \theta_1 > \theta_2 \) and \( T_1 > T_2 \).

### Final Answer:
- The correct conclusion is that \( \theta_1 > \theta_2 \).