Throw a ball straight up and catch it on the way down, at the same height. Taking into account air resistance, does the ball take longer to go up or to come down? Why?

The principle of free energy describes how non-equilibrium steady-states in living and non-living systems are maintained by limiting the number of states to a minimum. It proves that systems minimise a free energy function of their internal states, implying that hidden environmental states exists.

The best way to look at this problem is through the Energy Principle. Note that air resistance is a non-conservative force that continuously makes the total energy of the system to decrease. Also, note that the total energy of the system at any given point of the ball's trajectory is given by$E=K+U_g$.

Consider two points of the trajectory at the same altitude, one when the ball is rising and one when the ball is falling.

Write the equation for energy.

$$\begin{gathered} E_{\text{falling}}=E_{\text{rising}}-\left|W_{\text{air}}\right| \\ {U}_{\text{g, falling}}+K_{\text{falling}}=U_{\text{g, rising}}+K_{\text{rising}}-\left|W_{air}\right| \end{gathered}$$

As, the points at the same altitude so, $U_{\text{g, falling}}=U_{\text{g, rising}}$. Use this into the last obtained equation.

$$\begin{gathered} U_{\text{g, rising}}+K_{\text{falling}}=U_{\text{g, rising}}+K_{\text{rising}}-\left|W_{\text{air}}\right| \\ {K}_{\text{falling}}=K_{\text{rising}}-\left|W_{\text{air}}\right| \\ \frac{1}{2}m{v_{\text{falling}}}^2=\frac{1}{2}m{v_{\text{rising}}}^2-\left|W_{\text{air}}\right| \\ {v_{\text{falling}}}^2={v_{\text{rising}}}^2-\frac{2}{m}\left|W_{\text{air}}\right| \end{gathered}$$

From this it is concluded that, $v_{\text{rising}}>v_{\text{falling}}$.

This result implies that for any altitude, the ball rises faster than it falls. Since the ball covers the same distance in both directions upwards and downwards. So, it is concluded that the ball takes longer to come down.

Therefore, it is concluded that it takes more time for the ball to come down.