A body is projected up with velocity u. It reaches a point in its path at times `t_1` and `t_2` seconds from the time of projection. Then `(t_1+t_2)` is

Options

A. `(2u)/g`
B. `u/g`
C. `sqrt((2u)/g)`
D. `sqrt(u/g)`

Answer

To solve the problem, we need to analyze the motion of a body projected upwards with an initial velocity \( u \). The body reaches a certain height at two different times: \( t_1 \) while going up and \( t_2 \) while coming down. We are required to find the sum \( t_1 + t_2 \).

### Step-by-Step Solution:

1. **Understanding the Motion**:
- When a body is projected upwards, it moves against gravity until it reaches its maximum height. After reaching the maximum height, it falls back down to the original position.
- At a particular height, the body will take time \( t_1 \) to reach that height while going up and time \( t_2 \) to return to that height while coming down.

2. **Using Symmetry**:
- The motion of the body is symmetrical. This means that the time taken to go up to a certain height is equal to the time taken to come down from that height.
- Therefore, \( t_1 \) (time going up) and \( t_2 \) (time coming down) are related by the symmetry of the projectile motion.

3. **Total Time of Flight**:
- The total time of flight \( T \) for a body projected upwards with an initial velocity \( u \) can be calculated using the formula:
\[ T = \frac{2u}{g} \]
- Here, \( g \) is the acceleration due to gravity.

4. **Relating \( t_1 \) and \( t_2 \) to Total Time**:
- Since \( t_1 \) is the time taken to reach a certain height and \( t_2 \) is the time taken to return to the same height, we can express the total time of flight as:
\[ t_1 + t_2 = T \]
- Thus, we have:
\[ t_1 + t_2 = \frac{2u}{g} \]

5. **Conclusion**:
- Therefore, the sum \( t_1 + t_2 \) is equal to the total time of flight:
\[ t_1 + t_2 = \frac{2u}{g} \]

### Final Answer:
The value of \( t_1 + t_2 \) is \( \frac{2u}{g} \).