In an imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle's motion. A particle of mass m projected upward takes time `t_1` and reaching the maximum height and `t_2` in the return journey to the original point. Then

Options

A. `t_1ltt_2`
B. `t_1gtt_2`
C. `t_1=t_2`
D. the relation between `t_1 and t_2` depends n the mass of the particle.

Answer

To solve the problem, we need to analyze the motion of a particle projected upward in an imaginary atmosphere where a force \( F \) acts in the direction of the particle's motion. We will determine the time taken to reach the maximum height (\( t_1 \)) and the time taken to return to the original point (\( t_2 \)).

### Step-by-Step Solution:

1. **Identify Forces Acting on the Particle**:
- When the particle is moving upward, the forces acting on it are:
- Gravitational force: \( mg \) (downward)
- Force \( F \) (upward)
- The net force when moving upward is:
\[ F_{\text{net}} = F - mg \]
- This results in an upward acceleration \( a \):
\[ ma = F - mg \implies a = \frac{F - mg}{m} = \frac{F}{m} - g \]

2. **Using the Equation of Motion for Upward Journey**:
- The particle is projected with an initial velocity \( u \) and reaches the maximum height \( H \) in time \( t_1 \).
- At maximum height, the final velocity \( v = 0 \).
- Using the second equation of motion:
\[ H = ut_1 + \frac{1}{2} a t_1^2 \]
- Substituting \( a \):
\[ H = ut_1 + \frac{1}{2} \left(\frac{F}{m} - g\right) t_1^2 \]

3. **Finding the Time \( t_1 \)**:
- Rearranging the equation gives:
\[ 0 = ut_1 + \frac{1}{2} \left(\frac{F}{m} - g\right) t_1^2 - H \]
- This is a quadratic equation in \( t_1 \). Solving for \( t_1 \) leads to:
\[ t_1 = \sqrt{\frac{2H}{g - \frac{F}{m}}} \]

4. **Analyzing the Downward Journey**:
- When the particle is falling back down, the forces acting are:
- Gravitational force: \( mg \) (downward)
- Force \( F \) (also downward)
- The net force is:
\[ F_{\text{net}} = mg + F \]
- The downward acceleration \( a' \) is:
\[ ma' = mg + F \implies a' = g + \frac{F}{m} \]

5. **Using the Equation of Motion for Downward Journey**:
- The initial velocity at the start of the downward journey is \( 0 \) and the distance traveled is \( H \).
- Using the second equation of motion:
\[ H = 0 \cdot t_2 + \frac{1}{2} a' t_2^2 \]
- This simplifies to:
\[ H = \frac{1}{2} \left(g + \frac{F}{m}\right) t_2^2 \]

6. **Finding the Time \( t_2 \)**:
- Rearranging gives:
\[ t_2 = \sqrt{\frac{2H}{g + \frac{F}{m}}} \]

7. **Comparing \( t_1 \) and \( t_2 \)**:
- We have:
\[ t_1 = \sqrt{\frac{2H}{g - \frac{F}{m}}} \quad \text{and} \quad t_2 = \sqrt{\frac{2H}{g + \frac{F}{m}}} \]
- Since \( g - \frac{F}{m} < g + \frac{F}{m} \), it follows that:
\[ t_1 > t_2 \]

### Conclusion:
Thus, the time taken to reach the maximum height \( t_1 \) is greater than the time taken to return to the original point \( t_2 \):
\[ t_1 > t_2 \]