Question

a rubber ball dropped from the top of a ladder. It bounces in the same spot on the ground several times, each time to a lesser height that the previous times before resting

Answer 1

What you described is the usual bouncing-ball behavior controlled by the coefficient of restitution e (0 ≤ e ≤ 1), which gives the ratio of the rebound speed to the impact speed. Key points and formulas:

- If the ball is dropped from height H0, the impact speed is v = sqrt(2 g H0). After the bounce the upward speed is e v, so the height after the first bounce is
H1 = e^2 H0.
In general, the height after the n-th bounce is
Hn = e^{2n} H0.

- The ball in the idealized model makes infinitely many bounces but the heights form a geometric sequence tending to zero, so the motion converges to rest.

- Total vertical distance traveled (down + up) is finite:
S_total = H0 + 2 ∑_{n=1..∞} Hn = H0 (1 + e^2) / (1 − e^2) (for e < 1).

- Total time until (mathematically) infinite bounces converge:
t0 = sqrt(2 H0 / g) (time of first fall),
T_total = sqrt(2 H0 / g) · (1 + e) / (1 − e) (for e < 1).

- Energy after n bounces scales by e^{2n}: E_n = e^{2n} E0 (kinetic energy ∝ v^2).

- To estimate e from observation: e = sqrt(H1 / H0).

- To find how many bounces until the bounce height drops below some h_min:
n > ln(h_min / H0) / (2 ln e). (Use ceil/rounding appropriately; remember ln e < 0 so handle signs carefully.)

Practical note: real balls also lose energy to internal friction, sound, deformation and air resistance; for small heights these effects (and the discrete nature of internal dissipation) mean the ball actually stops after a finite number of visible bounces.

If you want, give H0 and either e or the measured first rebound height and I’ll compute the heights, total distance and time (or the number of visible bounces to a chosen threshold).