Why a bouncing ball never comes all the way back

Concept: each bounce keeps only a fraction e² of the height, because some energy turns to heat and sound.

● ball  ┈ drop height h₀  ┈ peak of each bounce  ▮ energy kept
e 0.80,
h₀ 2.00 m,
bounce # 0,
height 2.00 m,
v 0.0 m/s,
energy kept 100%
after a bounce  v′ = e·v  ⟹  hₙ = e²ⁿ·h₀;  energy ∝ h, so kept = e²ⁿ;  e = √(h_bounce / h_drop)

What you'll learn

The coefficient of restitution e. When the ball hits the floor it rebounds a little slower: the outgoing speed is e times the incoming speed (0 < e < 1). A perfect superball has e near 0.9; a lump of clay has e near 0.

Why the height drops as e²

Free-fall links speed and height by v² = 2gh, so height ∝ v². Since each bounce multiplies speed by e, it multiplies height by e². After n bounces the peak height is hₙ = e²ⁿ·h₀ — a geometric decay. The "lost" energy became heat, sound and deformation.

What the animation shows

The ball is integrated under gravity (g = 9.81 m/s²); at each floor contact its velocity is reversed and scaled by e. The orange dashes mark each successive peak (e²·previous) and the bar shows the fraction of the original energy still left (e²ⁿ).

Sources

Newtonian mechanics; Newton's experimental law of restitution (1687); v² = 2gh.