In an imaginary atmosphere, the air applies a small force F on a...

During the upward journey, the particle moves upward, so the force $F$ acts upward. The gravitational force $mg$ acts downward.

The net downward acceleration during ascent is $a_1 = g - \dfrac{F}{m}$.

Let the maximum height reached be $h$. The time of ascent $t_1$ is given by the kinematic equation $h = \dfrac{1}{2} a_1 t_1^2$, which gives $t_1 = \sqrt{\dfrac{2h}{a_1}}$.

During the downward journey, the particle moves downward, so the force $F$ acts downward. The gravitational force $mg$ also acts downward.

The net downward acceleration during descent is $a_2 = g + \dfrac{F}{m}$.

The time of descent $t_2$ for the same height $h$ is given by $h = \dfrac{1}{2} a_2 t_2^2$, which gives $t_2 = \sqrt{\dfrac{2h}{a_2}}$.

Since $F$ is a small positive force, we have $a_1 < a_2$.

Therefore, $\sqrt{\dfrac{2h}{a_1}} > \sqrt{\dfrac{2h}{a_2}}$, which implies $t_1 > t_2$.