Can you help show steps of how this problem really looks like and solve it?

Question

Suppose a ball is dropped from a heught of 6 meters. Using the formula h(n)=6(0.9)^1/5t n where h represents maximum height of the ball afterthe n+h bounce, what is the approximate percent rate of decrease?

Answer 1

Sure! The formula you typed out: h(n) = 6*(0.9)^(1/5)*n is the formula for the height of the ball after a certain number of bounces (n). The 6 in this formula is the height it is dropped from. We will call this h(0), because it is the height after 0 drops. You'll find that if you plug in 0 for n, you'll get h(0) = 6.

The percent decrease can be described like this: what percentage of the original height is missing after n bounces?

We can write this as 1- [h(n)/h(0)].

h(n)/h(0) gives us the percent OF the original height, so we subtract that from 100% to get us the missing height.

So, if we plug in h(0) instead of 6, we get:

h(n) = h(0)*(0.9)^(1/5)*n

If we divide both sides by h(0), we get:

h(0)/h(n) = 0.9^(1/5)*n

Which can be re-written as

h(0)/h(n) = (0.9^1/5)ⁿ ≈ 0.9812ⁿ

This is the percent of the original height. The percent decrease is 1-.9812 = 0.0188 = 1.88% decrease each bounce.