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

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: \[ \boxed{27 \, \text{m}} \]