Algebra 1 word problem

Question

two people are in need of escaping a bridge. the bridge is 800 feet tall. Ryan a bungee jumper who wants to jump from the bridge. Ryan's height above the ground at the top of each bounce is as follows: first bounce 200 feet off the ground; second bounce 80 feet above the ground; third bounce 32 feet above the ground. Sean is going to use a pulley system to lower himself off the bridge. at one pull he is 736 feet above the ground; at two pulls he is 672 feet above the ground; at three pulls he is 608 feet above the ground. write a function to model each jumper's descent.

Answer 1

Consider Sean:

The pattern is

Let n be the number of pulls.

Let h be the height.

h 800 736, 672, 608, ....

-64 -64 -64

n 0 1 2 3

Note: if we subtract 64 for each pull, we find his new height.

Because the change from each pull to the next is constant, think of constant slope. So a line would model his height.

Slope is change of one pull, so slope will be -64. Alternately, you can apply slope = (h₁- h₂) / (n₁- n₂) and also get 64.

y = mx + b or in our case

h = m *n + b

b is simply the initial value when n = 0. So Sean's height is

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h = -64 n + 800 feet.

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Ryan's height requires a bit more thought.

y 200 80 32

x 0 1 2 See ** below.

We look for a pattern, a math step to get from 800 to 200, 200 to 80 and so on.

How about multiplying each height by 4/10 which is the same as 2/5?

200 * 2/5 =80

80 * 2/5 = 32

So the pattern is multiply by 2/5 = 0.4

Repeated multiplication (200 initial * 2/5 * 2/5 calculates the third height) tells us to use an exponentiation model.

y = A ( R )^x

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y = 200 (2/5)^x feet

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One complicated, finicky detail.

** On Sean's example, the pattern worked out nicely for the initial bridge height of 800. This allowed us to use h = height before the pull and simplified finding b. However, for Ryan's jumps, 800 * 2/5 is not 200, so we could not use the 800 foot height of the bridge for Sean. There is no simple way to include the original 800 bridge height in Sean's model. Also, although using y = 200, when x =0 is counter-intuitive compared it very much simplifies figuring out the initial height I used.

My method is a bit of an oversimplification. The more thorough, but more complicated method is to create a model, leaving A as a variable. Then substitute in a point from the table. Solve for A.