Question

You drop a ball from a height of 0.5 meter. Each curved path has 52% of the height of the previous path. a. Write a rule for the sequence using centimeters. The initial height is given by the term n = 1. b. What height will the ball be at the top of the third path?

Answer 1

The initial height is .5 m. If the ball only reaches 52% of the previous max height, then after the third bounce you have:
.5 x (.52)^3=0.070304 meters as the height of the ball
☺☺☺☺

Answer 2

A rule for the sequence using centimeters is
`h_(n)=0.52^(n-1)*h_(1) \\`

and the height will the ball be at the top of the third path is 13.52 cm

Explanation:

You know that the initial height (h) is given by the term n = 1, then:

`h_(1)=0.5m=50cm`

`h_(2)=50cm*0.52`

52%=52/100=0.52

`h_(3)=50cm*0.52*0.52=50cm*0.52^2`

`h_(4)=50cm*0.52*0.52*0.52=50cm*0.52^3`

`h_(5)=50cm*0.52*0,52*0.52*0.52=50cm*0.52^4`

Then the standard equation is:

`h_(n)=0.52^(n-1)*h_(1) \\`

Now you need to calculate
`h_(3)`

`h_(3)=0.52^(3-1)*h_(1)=0.52^2*(50cm)=13.52 cm`