Question

A ball is dropped from a height of 200 meters. After striking the floor it re-bounces to \(\frac 45\)^th of the height from where it fell. The total distance it travels before coming to rest is ______ .

• A. 1200 meters
• B. 1600 meters
• C. 1800 meters
• D. 1820 meters

Answer 1

The Correct Option is C

The problem requires calculating the total distance traveled by a ball dropped from a given height before coming to rest. The ball falls from a certain height, hits the ground, rebounds to a fraction of its original height, and repeats this process until it stops bouncing.

Here's how to solve it:

1. The ball's initial drop height is 200 meters. 
2. Upon hitting the ground, the ball rebounds to \(\frac{4}{5}\) of the height from which it fell.

To find the total distance the ball travels, consider:

• The ball first falls 200 meters to the ground.
• It rebounds to \(\frac{4}{5} \times 200 = 160\) meters, then falls back another 160 meters.
• This makes each bounce effectively contribute twice its rebounded height to the total distance (once going up, once coming down).

We need to calculate the distance over an infinite number of bounces:

Let \(S\) be the total distance traveled:

\(S = 200 + 2 \times 160 + 2 \times 160 \left(\frac{4}{5}\right) + 2 \times 160 \left(\frac{4}{5}\right)^2 + \ldots\)

This sequence is a geometric progression with:

• First term \(a = 2 \times 160\)
• Common ratio \(r = \frac{4}{5}\)

The sum of an infinite geometric series is given by:

\(S = \frac{a}{1-r}\)

Substituting the values,

\(S = \frac{2 \times 160}{1-\frac{4}{5}}\)

\(S = \frac{320}{\frac{1}{5}}\)

\(S = 320 \times 5 = 1600\)

Adding the initial 200 meters drop, the total distance is:

\(200 + 1600 = 1800 \text{ meters}\)

Therefore, the total distance traveled by the ball before coming to rest is 1800 meters.