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Question 6 : A marble is dropped from a height of 3 metres onto the ground. After the hitting the ground, it bounces and reaches 80% of the height from which it was dropped. This repeats multiple times. Each time it bounces, the marble reaches 80% of the height previously reached. Eventually, the marble comes to rest on the ground.

What is the maximum distance that the marble travels from the time it was dropped until it comes to rest?

- 15 m
- 12 m
- 27 m
- 30 m
- 24 m

27 m

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Given that a marble is dropped from a height of 3 meters and every time it reaches up to 80% of height from where it is dropped.

For the first part of the motion it falls from a height of 3m and from here on everytime it bounces to a certain height it falls the same length and thus the distances appear twice in the summation

So the series is

3 + 3(0.8) + 3(0.8) + 3\((0.8)^{2}\) + 3\((0.8)^{2}\)+ 3\((0.8)^{3}\) + 3\((0.8)^{3}\) + . . . . . . . . .. .

3 + 2 [ 3(0.8) + 3\((0.8)^{2}\) + 3\((0.8)^{3}\) + . . . . . . ]

= 3 + 2[ \(\frac{3(0.8)}{1-\frac{4}{5}}\)] { \(S_{n}\) = \(\frac{a}{1-r}\)}

= 3 + 24

= 27

Therefore, the answer should be 27m.

Choice C is the correct answer.

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