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Worked solution

Count travelled legs in a bouncing-ball model

A student drops a ball made of a new compound from 200200 cm onto a hard surface. Each rebound reaches 89\frac89 of the previous height.

(i) Find the distance travelled when it reaches the highest point after the fourth bounce, to the nearest centimetre.

(ii) The ball is considered to have stopped when a bounce first reaches a height below 0.010.01 cm. Find the number of bounces and the total distance travelled, to the nearest centimetre.

[2017/SP/I/4]

CJC · Sequences and Series · Example 35 (Kinematics)

(i) Distance to the fourth rebound apex

Answer

12771277 cm (nearest cm).

Draw the legs and identify the height pattern

hk=200qk,q=89,T4=200+2h1+2h2+2h3+h4h_k=200q^k,\quad q=\frac89,\qquad T_4=200+2h_1+2h_2+2h_3+h_4

The source callout recommends a diagram, a table and observing the pattern. Put q=8/9. The initial descent is 200 cm, rebounds 1–3 each contribute an ascent and a descent, and rebound 4 contributes only its ascent.

Substitute before rounding

T4=200+400(89)+400(89)2+400(89)3+200(89)4≈1277.396738 cm≈1277 cm\begin{aligned}&T_4\\ & =200+400\left(\frac89\right)\\ &{}+400\left(\frac89\right)^2+400\left(\frac89\right)^3\\ &{}+200\left(\frac89\right)^4\approx1277.396738\ \text{cm}\approx1277\ \text{cm}\end{aligned}

Keep the fourth height singly rather than doubling every rebound.

How did this part go?

(ii) Stopping threshold and distance

Answer

8585 bounces; 34003400 cm (nearest cm).

Solve the strict height inequality

200(89)n<0.01,(89)n<0.01200,n>ln⁡(0.01/200)ln⁡(8/9)≈84.082461,nmin⁡=85.\begin{aligned}200\left(\frac89\right)^n&<0.01,\\\left(\frac89\right)^n&<\frac{0.01}{200},\\n&>\frac{\ln(0.01/200)}{\ln(8/9)}\approx84.082461,\\n_{\min}&=85.\end{aligned}

The logarithm of 8/9 is negative, so dividing reverses the inequality. The source cutoff is displayed as 84.082.

Sum the completed up-and-down rebounds

Tfloor,85=200+400∑k=184(89)k=200+400(8/9)[1−(8/9)84]1−8/9≈3399.838438 cm.\begin{aligned}T_{\mathrm{floor},85}&=200+400\sum_{k=1}^{84}\left(\frac89\right)^k\\&=200+400\frac{(8/9)[1-(8/9)^{84}]}{1-8/9}\\&\approx3399.838438\ \text{cm}.\end{aligned}

The source sums 84 paired rebound heights; after the 85th floor contact these completed legs have been travelled. The GP inside the bracket has first term 8/9, ratio 8/9 and 84 terms.

Include the tiny final ascent when ending at its apex

h85=200(8/9)85≈0.008976 cm,Tapex,85=Tfloor,85+h85≈3399.847414 cm,Tfloor,85≈Tapex,85≈3400 cmto nearest cm.\begin{aligned}h_{85}&=200(8/9)^{85}\approx0.008976\ \text{cm},\\T_{\mathrm{apex},85}&=T_{\mathrm{floor},85}+h_{85}\approx3399.847414\ \text{cm},\\T_{\mathrm{floor},85}&\approx T_{\mathrm{apex},85}\approx3400\ \text{cm}\quad\text{to nearest cm}.\end{aligned}

The lecturer notes that h85 is negligible at nearest-centimetre precision. Its comparison screen shows both totals: the floor-contact total and the apex total. Retain that rounding justification rather than treating the omitted ascent as exactly zero.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

An initial 200 cm descent, then three rebounds each travelled up and down, followed by only the upward leg of rebound 4. Rebound k has height 200(8/9)^k cm. Total to that apex is 200+2h1+2h2+2h3+h4.
An initial 200 cm descent, then three rebounds each travelled up and down, followed by only the upward leg of rebound 4. Rebound k has height 200(8/9)^k cm. Total to that apex is 200+2h1+2h2+2h3+h4.
An initial 200 cm descent, then three rebounds each travelled up and down, followed by only the upward leg of rebound 4. Rebound k has height 200(8/9)^k cm. Total to that apex is 200+2h1+2h2+2h3+h4.

Solution notes

The source table labels its heights in metres, but the question and computations use centimetres. It also describes the top-to-floor distance: the same height is travelled upwards after each contact. Here h_k consistently denotes the rebound height in cm. The source’s examiner-report note recommends initial height plus twice the other completed heights; the last endpoint still determines whether a final height is counted once.