Question

A rubber bouncy ball is dropped from a height of 119.00 inches onto a hard flat floor. After each bounce, the ball returns to a height that is 18.2% less than the previous maximum height. What is the maximum height reached after the 19th bounce?

What is the rate of change, b, for this situation?

Answer 1

Here the expression yields into geometric progression

Where

• First term=a=119
• Common ratio =100-0.182=0.818

Maximum height of 19th bounce

• a_n=ar^{n-1}
• a_n=119(0.818)¹⁸
• a_19=3.199ft

Rate of change is 0.818

Answer 2

• See below

Explanation:

18.2% less than h is:

• h*(100 - 18.2)/100 =
• h*0.818

The following series reflects this situation:

• 119, 119*0.818, 119*0.818², ...

This is a geometric series with the first term 119 and common ratio 0.818.

The maximum height after 19th bounce is:

• h = 119*0.818¹⁹ = 2.62 inches (rounded)

The rate of change is same as common ratio b = 0.818.