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

Step 1: Use the direct total-distance formula for a bouncing ball.
For a ball dropped from height $h$ that rebounds to a fraction $r$ of the height each time, the total distance it covers before coming to rest is $h \times \frac{1+r}{1-r}$. This bundles the first drop and every up-down bounce into one expression, so we don't need to sum the series term by term.

Step 2: Identify the values.
Here $h = 200$ m and $r = \frac{4}{5}$, since the ball always rebounds to four-fifths of the height it fell from.

Step 3: Plug into the formula.
Total distance $= 200 \times \frac{1 + \frac{4}{5}}{1 - \frac{4}{5}} = 200 \times \frac{\frac{9}{5}}{\frac{1}{5}} = 200 \times 9 = 1800$ m.

Step 4: Sanity check.
This matches summing the series directly: the first fall of 200 m plus twice the sum of $200 \times (4/5)^n$ for $n = 1,2,3,\ldots$, which also gives 1800 m.

Final Answer:
The ball travels 1800 m before it comes to rest. \[ \boxed{1800 \text{ m}} \]