Question

A ball is dropped from a height of 1 meter. Every time the ball bounces up, it reaches 50% of the height of the previous bounce (i.e., it rises up to 0.5 meters on the first bounce, 0.25 meters on the second bounce and so on). After an infinitely long time, the total distance covered by the ball in m is ...........

Hint:

For a bouncing ball that reduces its bounce height by a constant percentage, the total distance covered can be found using the sum of an infinite geometric series.

Answer 1

Correct Answer: 2.9

Step 1: Calculate the total distance covered.
The ball first falls 1 meter. After that, it bounces up to 0.5 meters, falls back down the same distance, and then bounces up to 0.25 meters, and so on. This creates an infinite geometric series for the distances covered after the initial fall.
The total distance is the sum of the series: \[ \text{Total distance} = 1 + 2 \times \left(0.5 + 0.25 + 0.125 + \dots \right). \] This is a geometric series with the first term \( a = 0.5 \) and the common ratio \( r = 0.5 \). The sum of an infinite geometric series is: \[ S = \frac{a}{1 - r} = \frac{0.5}{1 - 0.5} = 1. \] Thus, the total distance covered is: \[ \text{Total distance} = 1 + 2 \times 1 = 2 \, \text{meters}. \]

Step 2: Conclusion.
The total distance covered by the ball is 2 meters.