Pre- calc problem: A ball is dropped from a height of 30 feet.?

Question

Each time it strikes the ground, it bounces up 80% of the previous height. What total distance does the ball travel before it stops bouncing?

Answer 1

During the ball's initial fall, it travels 30 feet. It bounces, traveling up 30*.8 feet and down 30*.8 feet, for a grand total of 60*.8 feet on the first bounce. It then travels 60*.8² feet on the second bounce, 60*.8³ feet on the third bounce, and so on. All told, it travels:

(30 + 60*[k=1, ∞]∑.8^k) ft

This is a geometric series whose first term is .8 and whose common ratio is also .8 . There is a closed-form expression for the sum of such series, which is a/(1-r), where a is the first term and r is the ratio (obviously, this is only valid if the series actually converges, so |r| must be less than 1). So using this expression:

(30 + 60*.8/(1-.8)) ft

Writing the .8 as a fraction:

(30 + 60*(4/5)/(1-4/5)) ft

Simplifying:

(30 + 60*(4/5)/(1/5)) ft
(30 + 60*4) ft
(30+240) ft
270 ft

And we are done.

Answer 2

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Answer 3

The sum of half of the bounces (AFTER the initial drop, in which it traveled 30 feet) can be produced by an infinite geometric sum. The formula for this sum is equal to

the first term / (1 - common ratio)

The first bounce in this case is 24 since it's 30*80%

= 24 / (1-.80)
= 24/.2 = 120

So half the bounces total 120, therefore all the bounces are 240 + the original 30 = 270 feet.

Answer 4

It will bounce 270 feet.

Answer 5

37.5 aproxx

Answer 6

12.2 ft