A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way — each time rising to 60% of the previous height.

(A) What height does the ball reach after the $5^{\text{th}}$ bounce?

(B) What is the total vertical distance the ball has travelled by the time it hits the ground for the $6^{\text{th}}$ time?                                                                                                                                                                                                                       [Page No. 193]

Heights after each bounce form a GP with $a = 80 \times 0.6 = 48$ m (the 1st-bounce height) and $r = 0.6$.

$$(A) \text{ Bounce 1: } 80 \times 0.6 = 48 \text{ m}$$

$$\text{Bounce 2: } 48 \times 0.6 = 28.8 \text{ m}$$

$$\text{Bounce 3: } 28.8 \times 0.6 = 17.28 \text{ m}$$

$$\text{Bounce 4: } 17.28 \times 0.6 = 10.368 \text{ m}$$

$$\text{Bounce 5: } 10.368 \times 0.6 = 6.2208 \text{ m}$$

(B) The ball hits the ground the 6th time means it has fallen 6 times and bounced up 5 times (after the 6th hit it has not yet bounced).

Down distances: 80, 48, 28.8, 17.28, 10.368, 6.2208 (six fall distances).

Up distances: 48, 28.8, 17.28, 10.368, 6.2208 (five rise distances).

$$\begin{aligned} \text{Sum down} &= 80 + 48 + 28.8 + 17.28 + 10.368 + 6.2208 = 190.6688 \text{ m} \end{aligned}$$

$$\begin{aligned} \text{Sum up} &= 48 + 28.8 + 17.28 + 10.368 + 6.2208 = 110.6688 \text{ m} \end{aligned}$$

$$\begin{aligned} \text{Total distance} &= 190.6688 + 110.6688 \\ &= 301.3376 \text{ m} \approx 301.34 \text{ m} \end{aligned}$$