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The first part of the process is a ball in free flight from one step down to the next:

Consider gravitational potential energy $m g h$ (where m is mass, g is the force of gravity, and h is the height) at the peak height $H$ of the flight path over steps of height $S$ relative to A and B:

$$\begin{aligned} E_A &= m g (H - S) \\ E_B &= m g H \end{aligned}$$

Consider kinetic energy $\frac{1}{2} m v^2$ at A and B:

$$\begin{aligned} E_A &= \frac{1}{2} m v_A^2 \\ E_B &= \frac{1}{2} m v_B^2 \end{aligned}$$

Some simple algebra gives:

$$\begin{aligned} v_A = \sqrt{ 2 g (H - S) } \\ v_B = \sqrt{ 2 g H } \end{aligned}$$

When in flight, the only force acting on the ball is gravity. Using Newton's law of motion relating force and acceleration $F = m a$ and integrating with the initial conditions at $A$ with $t = 0$ gives:

$$\begin{aligned}\frac{\partial}{\partial t}\frac{\partial}{\partial t} y & = -g \\ \frac{\partial}{\partial t} y &= v_A - g t \\ y &= r + v_A t - \frac{1}{2} g t^2 \end{aligned}$$

where $y$ is the height of the ball center and $r$ is the ball radius. Now we can find the duration $T_f$ of the flight, which ends when $y(T_f) = y(0) - S$:

$$\begin{aligned} r - S &= r + v_A T_f - \frac{1}{2} g T_f^2 \\ \frac{1}{2} g T_f^2 - v_A T_f - S &= 0 \\ T_f = \frac{v_A + \sqrt{ v_A^2 + 2 g S } }{ g } \end{aligned}$$

The second part of the process is a compressive bounce when the ball reaches the lower step:

Modelling the bounce by a damped harmonic oscillator gives an equation of motion, where $d$ is the maximal squashing of the ball:

$$y = r \left( 1 - d \sin (\beta t) e^{-\gamma t} \right)$$

Here $\beta$ determines the speed of the bounce and $\gamma$ determines the energy lost during the bounce. Using the boundary conditions $v_C = v_B$ and $v_D = v_A$ for a smooth transition from free flight to bouncing, and setting $T_b$ to be the duration of the bounce (time of first return to initial $y$ value), we can differentiate the equation of motion and solve for the unknowns:

$$\begin{aligned} \frac{\partial}{\partial t} y &= r e^{-\gamma t} \left( \gamma \sin(\beta t) - d \beta \cos(\beta t) \right) \\ v_B &= r d \beta \\ \beta &= \frac{v_B}{r d} \\ T_b &= \frac{\pi}{\beta} \\ v_A &= r e^{-\gamma T_b} d \beta \\ \gamma &= -\frac{\log \frac{v_A}{r d \beta}}{T_b} \end{aligned}$$

The time taken for the whole loop is $T = T_f + T_b$. Assuming the ball is initially spherical, and that volume remains constant, we can compute the squashing factor using the property of an ellipsoid that $V = k r_x r_y r_z$ assuming that $r_x = r_z$:

$$\begin{aligned} r_y &= y \\ r_x &= \sqrt{\frac{r^3}{y}} \end{aligned}$$

Implementing all these equations in the Haskell programming language using the Diagrams library for visualization gives some cute animated GIFs:

You can download the full source code for this post.