Question

Function h describes the height of a ball, in inches, after n bounces and is defined by the equation h(n)=120×(4/5)^n.a. What is h(3)? What does it represent in this situation?b. Could h(n) be 150? Explain how you know?c. Which ball loses its height more quickly, this ball or a tennis ball whose height in inches after n bounces is modeled by the function f where f(n)=50×(5/9)^n?d. How many bounces would it take before the ball bounces less than 12 inches from the surface?

Answer 1

a.

When we find h(3), we found the height of a boll after 3 bounces.

Then:

$h(n)=120\cdot(\frac{4}{5})^n$$h(3)=120\cdot(\frac{4}{5})^{3^{}}$$h(3)=61.44$

Hence, after 3 bounces, the height of the ball is 61.44 in.

b) . Could h(n) be 150?

Let's set the equation equal to 150:

h(n)=150

$120\cdot(\frac{4}{5})^n=150$

Solve for n:

Divide both sides by 120:

$\frac{120\cdot(\frac{4}{5})^n}{120}=\frac{150}{120}$$(\frac{4}{5})^n=\frac{5}{4}$

Where n=-1

If n represents the number of bounces, then, it can be equal to a negative number. Hence, h(n) can not be equal to 150.

c.

The equation for this ball is h(n)=120×(4/5)^n.

The equation for a tennis ball is modeled by f(n)=50×(5/9)^n.

We can set some values for n and then look at the height for each ball:

For n=2

The given ball:

h(n)=120×(4/5)^n

h(2)=120×(4/5)^2=76.8

Tennis ball

f(n)=50×(5/9)^n

f(2)=50×(5/9)^2 = 15.43

For n = 3

The given ball:

h(n)=120×(4/5)^n

h(3)=120×(4/5)^3=61.44

Tennis ball

f(n)=50×(5/9)^n

f(3)=50×(5/9)^3 = 8.57

for n=4

The given ball:

h(n)=120×(4/5)^n

h(4)=120×(4/5)^4=49.152

Tennis ball

f(n)=50×(5/9)^n

f(4)=50×(5/9)^4 = 4.76

Looking at the n number of bonces. the tennis ball is losing around the half-height after each bonce.

Hence, the tennis ball loses its height more quickly

d.

Lets set f(n)<12 inches

Then:

$120\cdot(\frac{4}{5})^n<12$

Solve for n:

$\frac{120\cdot(\frac{4}{5})^n}{120}<\frac{12}{120}$$(\frac{4}{5})^n<\frac{1}{10}$

Then

$n<\log _{\frac{4}{3}}(\frac{1}{10})$$n<10.311$

It would take less than 10 bounces before the ball reaches the height of 12 inches.