# Geometric Series: Rebounding Ball — a Ball Rebounds

INFINITE GEOMETRIC SERIES IN REAL LIFE
Examples for when common ratios are Percentages

If you have ever bounced a ball, you know that when you drop it, it never rebounds to the height from which you dropped it. Suppose a ball is dropped from a height of three feet, and each time it falls, it rebounds to 60% of the height from which it fell. The heights of the ball’s rebounds form a sequence. Figure 1

GEOMETRIC SEQUENCES
The height of the first rebound of the ball is 3(0.6) or 1.8 feet. The height of the second rebound is 1.8(0.6) or 1.08 feet. The height of the third rebound is 1.08(0.6) or 0.648 feet. The sequence of heights, 1.8, 1.08, 0.648, …, is an example of a geometric sequence. A geometric sequence is a sequence in which each term after the first is found by multiplying the previous term by a constant r called the common ratio.

As with an arithmetic sequence, you can label the terms of a geometric sequence as a1, a2, a3, and so on. The nth term is an and the previous term is a(n-1). So, an=a(n-1)r. Thus, r=an/a(n-1) . That is, the common ratio can be found by dividing any term by its previous term.

GEOMETRIC PROGRESSION (G.P.)
Suppose a ball always rebounds exactly 70 percent of the distance it falls. For instance, if this ball falls from a height of 100 units, then it will rebound exactly 70 units. As a result the second fall will be from a height of 70 units. See Figure 2.

Now the ball will rebound exactly (0.7)⋅(70)=49 units and so, on. Figure 2

The following table gives each height through which the ball bounces in the first 5 falls. If you look carefully you will find that each term in sequence
100, 100(0.7), 100(0.7)2, 100(0.7)3, 100(0.7)4, …(except the first) is obtained by multiplying the previous term by a fixed constant 0.7.

Such a sequence is called geometric sequence or geometric progression or brieﬂy, G.P. In other words, a geometric sequence or geometric progression is a sequence in which each term, except the first, is obtained by multiplying the term immediately preceding it by a fixed, non-zero number. The fixed number is called the common ratio.

Examples for when common ratios are Fractions

Rebounding Ball
📌 Example 1. A certain ball when dropped from a height rebounds ⅗ of the original height. How high will the ball rebound after the fourth bounce if it was dropped from a height of 10 m?
✍ Solution: A diagram as to how the ball will rebound will give a clear view of the problem. The problem asking for how high the ball will rebound after the 4”‘ bounce is equivalent to asking for the height the ball has reached on its fourth rebound, then this can be solved by listing down the heights the ball reaches at each rebound. On the first rebound the height the ball reaches is ⅗ the height of the previous height, which is 10m.
So that got ⅗ of 10=⅗⋅10=6m. On the second rebound the height the ball reaches is ⅗⋅6=18/5; on the third rebound, the height is ⅗⋅18/5=54/25; and finally on the fourth rebound, the height the ball rebounds is ⅗⋅54/25=162/125=1.3 m.

Using the formula for the nth term of a geometric sequence with a1=6, and r=⅗: The ball rebounds 1.3 m after the 4th bounce.

📌 Example 2. What is the total distance that the ball in Example-1 has traveled by the time it comes to rest?
✍ Solution: If the ball always rebounds ⅗ the distance it fails, then theoretically it never comes to rest. However, the sum of the distances it travels downward and the sum of the distances it travels on the rebounds form two infinite series. The total distance S the ball travels can be found by adding the sums of these infinite series. This gives

S=downward series+upward series Notice that the series (ii) is contained in both the downward and upward series, then S can be written as Using the formula for the sum of an infinite series S=a1/(1-r) with a=6 and r=⅗, then, The total distance traveled by the ball is 40 m.

📌 Example 3: Follow a bouncing ball
Suppose a ball always rebounds ⅔ of the height from which it falls and the ball is dropped from a height of 6 feet. Find the total distance that the ball travels.
✍ Solution:
The ball falls 6 feet (ft) and rebounds 4 ft, then falls 4 ft and rebounds 8/3 ft. The following series gives the total distance that the ball falls: The distance that the ball rebounds is given by the following series: Each of these series is an infinite geometric series with ratio ⅔. Use the formula for an infinite geometric series to find each sum: The total distance traveled by the ball is the sum of F and R, 30 ft.
Let’s read a next post Geometric Series: Bouncing Balls — a Ball Bounces.