Question

Can someone please help me how to do this problem Algebra 2 xx Medal awarded!

1. A ball rebounds to a height of 30.0 cm on the third bounce and to a height of 5.2 cm on the sixth bounce.
a. Find two points and plug into the point-ratio form y=y1*b^x-x1, using r for the ratio. Let x represent the bounce number, and let y represent the rebound height in centimeters.

b. Solve the equation for b. Plug b and one of the two points into the point*ratio form to get an equation
c.what height was the ball dropped from?

Answer 1

i am sure we can do this, but what the monkey is the "point - ratio" formula?

Answer 2

the point ratio formula is y=y1*b^x-x1

Answer 3

ok i see
i think it is
\[\large y=y_1b^{x-x_1}\] right?

Answer 4

yes yes :)

Answer 5

\[y=30b^{x-3}\] and
\[y=5.2b^{x-6}\] like that right?

Answer 6

lets see if we can solve for \(b\)

Answer 7

well one point is (x1,y1) and the other point goes in as (x,y)

Answer 8

I think the points are (30,3) and (5.2, 6)

Answer 9

\[30b^{x-3}=5.2b^{x-6}\] divide and get
\[b^{x-3}=\frac{5.2b^{x-6}}{30}\] divide by \(b^{x-6}\) and get
\[\frac{b^{x-3}}{b^{x-6}}=\frac{5.2}{30}\]

Answer 10

you have the x and y backwards

Answer 11

no matter, however, we are almost done

Answer 12

so the two points would have to be (3,30) and (6,5.2)?

Answer 13

\[\large\frac{b^{x-3}}{b^{x-6}}=b^{x-3-(x-6)}=b^3\]

Answer 14

yes, the number of bounces is the \(x\) and the height is the \(y\)

Answer 15

so finally we have
\[\large b^3=\frac{5.2}{30}\] making \[\large b=\sqrt[3]{\frac{5.2}{30}}\] whatever that is

Answer 16

i am sure there is an easier way to do this, but no matter

Answer 17

now you need a calculator to find \(b\)

Answer 18

looks like \(b\) is about \(0.5576\)
http://www.wolframalpha.com/input/?i=cubed+root+%285.2%2F30%29

Answer 19

@Mertsj you ever seen anything like this before?

Answer 20

oh I see.. since c is asking the height that the ball was dropped from would i have to plug that value in or..

Answer 21

ok so we have the answer to a and b right?

Answer 22

correct :)

Answer 23

now you can use
\[y=30\times (.5576)^{x-3}\]

Answer 24

if you want the height the ball was dropped from, that is the "zeroth" bounce or \(x=0\) plug (0\) in to \[y=30\times (.5576)^{x-3}\] get \[y=30\times (.5576)^{-3}\] then a calculator

Answer 25

i have to say this is a first for me

Answer 26

oh wow .. :) you are very smart though:-)) it is kinda of confusing tbh

Answer 27

@satellite73 You're too smart for your own good.

Answer 28

lol hardly
you ever seen this before?

Answer 29

usually you would say (i think) \(ar^3=30\) and \(ar^6=5.2\) then solve

Answer 30

solve for ar?

Answer 31

solve for \(a\) and \(r\) but not matter, we are done

Answer 32

*no matter

Answer 33

this method is much easier i think
\[\frac{ar^6}{ar^3}=\frac{5.2}{30}\] or
\[r^3=\frac{5.2}{30}\] so
\[r=\sqrt[3]{\frac{5.2}{30}}\] same as before exactly but i used \(r\) instead of \(b\) and solved quicker

Answer 34

Oh i see.. so just to clarify :) for c it would be ar^3=30 and ar^b=5.2?

Answer 35

forget what i just wrote
we already solved for \(b\) and we get c via \[y=30\times (.5576)^{-3}\]

Answer 36

and a calculator

Answer 37

http://www.wolframalpha.com/input/?i=30%28.5576%29^%28-3%29
looks like about 173

Answer 38

ok thanks again so much you've been really very helpful! :)) xx