A ball is thrown vertically upwards from the ground. If `T_1` and `T_2` are the respective time taken in going up and coming down, and the air resistance is not ignored, then

Options

A. `T_1gtT_2`
B. `T_1=T_2`
C. `T_1 lt T_2`
D. nothing can be said

Explanation

To solve the problem, we need to analyze the motion of the ball as it is thrown upwards and then comes back down, taking into account the effects of air resistance.

### Step-by-Step Solution:

1. **Identify Forces Acting on the Ball:**
- When the ball is thrown upwards, two forces act on it:
- The gravitational force (\(mg\)) acting downwards.
- The air resistance force (\(F_1\)) acting downwards as well (since the ball is moving upwards).
- Therefore, the net force acting on the ball while it is going up is:
\[ F_{\text{net, up}} = mg + F_1 \]

2. **Write the Equation of Motion for Going Up:**
- According to Newton's second law, the net force is equal to mass times acceleration (\(ma\)):
\[ mg + F_1 = ma_1 \]
- Rearranging this gives us the acceleration while going up:
\[ a_1 = \frac{mg + F_1}{m} \]

3. **Identify Forces Acting on the Ball When Coming Down:**
- When the ball is coming down, the forces acting on it are:
- The gravitational force (\(mg\)) acting downwards.
- The air resistance force (\(F_2\)) acting upwards (since the ball is moving downwards).
- The net force acting on the ball while it is coming down is:
\[ F_{\text{net, down}} = mg - F_2 \]

4. **Write the Equation of Motion for Coming Down:**
- Again, using Newton's second law:
\[ mg - F_2 = ma_2 \]
- Rearranging gives us the acceleration while coming down:
\[ a_2 = \frac{mg - F_2}{m} \]

5. **Compare the Accelerations:**
- From the equations derived:
- \(a_1 = \frac{mg + F_1}{m}\)
- \(a_2 = \frac{mg - F_2}{m}\)
- Since \(F_1\) (air resistance while going up) is positive and \(F_2\) (air resistance while coming down) is also positive, we can conclude:
\[ a_1 > a_2 \]
- This indicates that the acceleration while going up is greater than the acceleration while coming down.

6. **Relate Time Taken to Accelerations:**
- The time taken to go up (\(T_1\)) and the time taken to come down (\(T_2\)) are inversely related to their respective accelerations:
\[ T_1 \propto \frac{1}{a_1} \quad \text{and} \quad T_2 \propto \frac{1}{a_2} \]
- Since \(a_1 > a_2\), it follows that:
\[ T_1 < T_2 \]

### Conclusion:
The time taken to go up (\(T_1\)) is less than the time taken to come down (\(T_2\)). Therefore, the correct answer is:
\[ T_1 < T_2 \]