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.
- (i) What height does the ball reach after the 5th bounce?
- (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
Step-by-step solution
Idea: Bounce heights form a GP: 48, 28.8, … with r = 0.6. Before the 6th landing the ball falls 80 m once, then goes up and comes down again for each of the first 5 bounces.
(i) What height does the ball reach after the 5th bounce?
- Bounce heights: 80 × 0.6 = 48, 28.8, 17.28, 10.368, … a GP with a = 48, r = 0.6.½ mark
- 5th bounce = 80 × 0.65 = 80 × 0.07776 = 6.2208 m.½ mark
(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?
- 1st landing: after falling 80 m. Each of the 5 bounces before the 6th landing goes up and down the same height.1 mark
- Sum of the 5 bounce heights = 48 + 28.8 + 17.28 + 10.368 + 6.2208 = 110.6688 m.1 mark
- Total = 80 + 2 × 110.6688 = 80 + 221.3376 = 301.3376 m (≈ 301.34 m).1 mark
Check: Landings: 1st after the 80 m drop, then one more landing after each bounce: bounces 1 to 5 give landings 2 to 6 ✓.
Answer to write in the exam
(i)
Bounce heights: GP, a = 80 × 0.6 = 48, r = 0.6
5th bounce = 80 × 0.65 = 80 × 0.07776
∴ Height after 5th bounce = 6.2208 m
(ii)
Bounce heights: 48, 28.8, 17.28, 10.368, 6.2208 m
Sum = 110.6688 m
Total distance = 80 + 2 × 110.6688 = 80 + 221.3376
∴ Total distance = 301.3376 m ≈ 301.34 m
Common mistakes that cost marks
- Using 80 × 0.64 for the 5th bounce (counting the drop as a bounce).
- Counting each bounce height only once. The ball goes up and comes back down, so each bounce height is travelled twice.
- Including the 6th bounce’s rise. The question stops when the ball hits the ground for the 6th time.
How this can come in the exam
A ball dropped from 100 m rises to 50% of its height after each bounce. The height after the 3rd bounce is
- 25 m
- 12.5 m
- 6.25 m
- 50 m
Show answer
(B) 12.5 m
50, 25, 12.5.
In a science lab a ball is dropped from 10 m onto a hard floor and each bounce reaches 80% of the previous height.
(i) Write the heights of the first three bounces. (ii) Are these heights in AP or GP? Give the common ratio or difference. (iii) Find the total distance travelled until the ball hits the floor for the 3rd time.
Show answer
(i) 8 m, 6.4 m, 5.12 m (1 mark). (ii) GP with common ratio 0.8 (1 mark). (iii) 10 + 2 × (8 + 6.4) = 10 + 28.8 = 38.8 m (2 marks).Try one yourself
A ball is dropped from 125 m and rises to 40% of the previous height each bounce. Find the height of the 3rd bounce and the distance travelled until the 3rd landing.
Show answer
Bounces: 50, 20, 8 m. Distance until 3rd landing = 125 + 2 × (50 + 20) = 265 m.
More questions like this
- Which term of the sequence 2, 2√2, 4, … is 128?
- The figure shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on.
Look at the figure and try to answer the following questions. - Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
- Determine the AP whose third term is 16 and whose 7th term exceeds the 5th term by 12.
- How many three-digit numbers are divisible by 7?
(Hint: All three-digit numbers divisible by 7 form an AP. Find the smallest and largest such three-digit numbers.)
All Sequences and progressions questions · All maths questions