Question

A ball is dropped from a height of 200 meters. After striking the floor it rebounces to \(\frac{4}{5}\)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

Hint:

Add the first fall to twice the infinite geometric series of the rebound heights, using ratio 4/5.

Answer 1

The Correct Option is C

To solve the problem of determining the total distance a ball travels before coming to rest after being dropped from a height of 200 meters and rebounding to \(\frac{4}{5}\) of its previous height, follow these steps: 

1. The ball is initially dropped from a height of 200 meters. Thus, it first travels 200 meters downward to the ground.
2. After striking the ground, the ball rebounds to a height of \(\frac{4}{5}\) of 200 meters.

\(Rebound\ Height = \frac{4}{5} \times 200 = 160\ meters\)

1. The ball then falls again from this 160-meter height, adding further to the total distance.
   • It travels 160 meters downward.
   • It rebounds to \(\frac{4}{5}\) of 160 meters.

\(= \frac{4}{5} \times 160 = 128\ meters\)

1. This sequence of downward and upward travels continues indefinitely, forming an infinite geometric series.

The total distance the ball travels can be calculated using the formula for the sum of an infinite geometric series:

• The sum of the distances traveled in the upward direction can be calculated using the series formula for the upward distances:

\(S_1 = 160\ +\ 128\ +\ 102.4\ +\ ...\)
Where a = 160 meters and r = \(\frac{4}{5}\)
\(S_1 = \frac{160}{1 - \frac{4}{5}} = 800\ meters\)

• Similarly, the sum of the distances in the downward direction, excluding the initial drop:

\(S_2 = 160\ +\ 128\ +\ 102.4\ +\ ...\)
\(S_2 = \frac{160}{1 - \frac{4}{5}} = 800\ meters\)

Hence, the total distance covered by the ball, including the initial drop of 200 meters, is:

\(Total\ Distance = 200\ +\ 800\ +\ 800\ = 1800\ meters\)

Therefore, the total distance the ball travels before coming to rest is 1800 meters. Thus, the correct answer is: 1800 meters.

Answer 2

Step 1: Note the first fall.
The ball falls the first time from 200 m, so it covers 200 m before it hits the floor for the first time.

Step 2: Model each bounce as an up-and-down trip.
After hitting the floor, the ball rises to \(\frac{4}{5}\) of 200 m, then falls back the same distance, so that bounce contributes \(2 \times \frac{4}{5} \times 200\) m. The next bounce rises to \(\frac{4}{5}\) of that height, contributing \(2 \times \left(\frac{4}{5}\right)^2 \times 200\) m, and so on forever.

Step 3: Add up all the bounces as an infinite GP.
Total bounce distance \(= 2 \times 200 \times \left(\frac{4}{5} + \left(\frac{4}{5}\right)^2 + \left(\frac{4}{5}\right)^3 + \cdots \right)\). This is an infinite GP with first term \(\frac{4}{5}\) and common ratio \(\frac{4}{5}\), so its sum is \(\frac{4/5}{1 - 4/5} = \frac{4/5}{1/5} = 4\).

Step 4: Compute the total distance.
Total bounce distance \(= 400 \times 4 = 1600\) m. Add the first fall of 200 m: total distance \(= 200 + 1600 = 1800\) m.

Final Answer:
The ball travels 1800 m in total before coming to rest. \[ \boxed{1800 \text{ m}} \]