The bounce heights form a GP with first term 48 metres and common ratio 0.6. Therefore, the fifth bounce height is 48 multiplied by 0.6 raised to 4, which equals 6.2208 metres. Hence, the required height is 6.2208 metres.
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?
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(i) Answer:
After the first bounce, the ball reaches 60% of 80 metres, which is 48 metres. The successive bounce heights form a GP with first term 48 and common ratio 0.6. Therefore, the height after the fifth bounce is 48 multiplied by 0.6 raised to 4. This equals 6.2208 metres. Hence, the ball reaches a height of 6.2208 metres after the fifth bounce.
(ii) Answer:
The ball first travels 80 metres downward. Before hitting the ground for the sixth time, it rises and falls five times. The five heights are 48, 28.8, 17.28, 10.368 and 6.2208 metres. Their sum is 110.6688 metres. Therefore, total vertical distance equals 80 plus twice 110.6688, which is 301.3376 metres. Hence, the ball has travelled 301.3376 metres when it hits the ground for the sixth time.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 8 Predicting What Comes Next: Exploring Sequences and Progressions Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-8/