Question

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?

• A. 15 m
• B. 12 m
• C. 24 m
• D. 27 m
• E. 30 m

Answer 1

The Correct Option is D

To find the maximum distance the marble travels, we need to consider both the downward and upward paths of the marble.

The marble is initially dropped from a height of 3 meters. After hitting the ground, it bounces back to 80% of its previous height.

1. Initial fall (downward): 3 meters
2. The marble bounces back to 80% of 3 meters: \(3 \times 0.8 = 2.4\) meters (upwards).

The process of bouncing continues with each subsequent height being 80% of the previous bounce height. Let's calculate both up and down distances for each bounce until the motion effectively stops.

For the second bounce:

• Falls from 2.4 meters: 2.4 meters (downward)
• Bounces back to 80% of 2.4 meters: \(2.4 \times 0.8 = 1.92\) meters (upwards)

For the third bounce:

• Falls from 1.92 meters: 1.92 meters (downward)
• Bounces back to 80% of 1.92 meters: \(1.92 \times 0.8 = 1.536\) meters (upwards)

This sequence forms a geometric series. The distances for reviews up and down are equal for the bounce, except for the first fall where no upward bounce contributes.

Sum of downward travels: \(3 + 2.4 + 1.92 + \ldots\) 
This forms a geometric series with first term \(a = 3\) and common ratio \(r = 0.8\).

Sum of infinite geometric series formula: \(S = \frac{a}{1 - r}\)

Calculating the downward path:

• Total downward distance: \(\frac{3}{1 - 0.8} = \frac{3}{0.2} = 15\) meters

Sum of upward paths: First upward bounce starts from 2.4 meters: \(2.4 + 1.92 + 1.536 + \ldots\)

For upward series, the first term \(a = 2.4\) and the common ratio \(r = 0.8\)

• Total upward distance: \(\frac{2.4}{1 - 0.8} = \frac{2.4}{0.2} = 12\) meters

The total distance travelled by the marble is the sum of both downward and upward distances.

The total distance is: \(15 + 12 = 27\) meters

Thus, the maximum distance that the marble travels is 27 meters.

Answer 2

A marble is dropped from a height of \(3 \, \text{m}\). Each time it hits the ground, it bounces back to \(80\%\) of the previous height.

Step 1: First fall

The marble first falls from the height of \[ 3 \, \text{m} \]

Step 2: First bounce

The marble bounces back to \[ 0.8 \times 3 = 2.4 \, \text{m} \] After reaching that height, it again falls \(2.4 \, \text{m}\).

Step 3: Subsequent bounces

Each time, the marble covers two segments: going up and then coming down. Heights form a geometric progression (GP): \[ 2.4, \; 1.92, \; 1.536, \; \dots \] with first term \(a = 2.4\) and common ratio \(r = 0.8\).

Step 4: Total distance

Total distance travelled by the marble: \[ \text{Distance} = \text{First fall} + 2 \times (\text{Sum of GP}) \]

Sum of the infinite GP: \[ S = \frac{a}{1-r} = \frac{2.4}{1-0.8} = \frac{2.4}{0.2} = 12 \]

Hence total distance: \[ \text{Distance} = 3 + 2 \times 12 = 3 + 24 = 27 \, \text{m} \]

Final Answer

The maximum distance the marble travels until it comes to rest is: \[ \boxed{27 \, \text{m}} \]