A Geometric Series Problem: The Bouncing Ball

The bouncing ball is a classic applied geometric progression: the rebound heights form a GP with ratio $< 1$, so their sum to infinity is finite. The one subtlety is that each rebound is travelled twice (up and down), while the first drop happens once.

A ball is dropped from a height of $10$ m. Each time it hits the ground it rebounds to $0.6$ of its previous height. Find the total distance the ball travels before coming to rest. [5]

The working

Step 1, separate the first drop from the rebounds. The ball first falls $10$ m (once). After that, each rebound height is travelled up and down.

The rebound heights form a GP: first rebound $10 \times 0.6 = 6$, then $6 \times 0.6$, and so on, first term $a = 6$, ratio $r = 0.6$.

Step 2, sum the rebound heights to infinity (valid since $|r| = 0.6 < 1$): $S_\infty = \frac{a}{1 - r} = \frac{6}{1 - 0.6} = \frac{6}{0.4} = 15 \text{ m} \quad \text{(M1, A1)}$

Step 3, account for up-and-down. Each rebound height is covered twice, so the total rebound distance is $2 \times 15 = 30$ m. (M1)

Step 4, add the initial drop: $\text{total distance} = 10 + 2(15) = 10 + 30 = 40 \text{ m} \quad \text{(A1)}$

Where the marks are won and lost

• The first drop is counted once, the rebounds twice. The formula is (initial drop) $+ 2 \times$ (sum of rebounds), not $2 \times$ everything.
• The rebound GP starts at $6$ (the first rebound), not $10$. Using $a = 10$ double-counts the initial height.
• $|r| = 0.6 < 1$ justifies the sum to infinity, worth noting.

Common mistakes

• Forgetting the factor of $2$ for up-and-down travel.
• Doubling the initial drop as well (it happens only once).
• Taking $a = 10$ instead of $6$ for the rebound series.

Full method: Sum to Infinity notes. See also Geometric Progressions. Topic home: Series pillar.