1. ## word problem

A ball dropped from a height of 4 ft above the ground, it bounces off the ground to a height that is 3/4 the height from which it was dropped,then falls again. The ball continues to bounce, and the height of each is 3/4 of the height proceeding bounce. Approximately what total vertical distance will the ball have traveled from the time it is dropped to the time it hits the ground the 5th time?
a. 14.5 ft b. 17.9ft c. 20.4 d. 22.3ft
I have 20.4ft.. Is that correct?

2. ## Re: word problem

$4 + 2(0.75)(4) + 2(0.75)^2(4) + 2(0.75)^3(4) + 2(0.75)^4(4) = 20.40625$

3. ## Re: word problem

.Hello, talanih!

A ball dropped from a height of 4 ft above the ground.
It bounces off the ground to a height that is 3/4 the height from which it was dropped, then falls again.
The ball continues to bounce, and the height of each is 3/4 of the height preceding bounce.
Approximately what total vertical distance will the ball have traveled from the time it is dropped
to the time it hits the ground the 5th time?

(a) 14.5 ft . . (b) 17.9 ft . . (c) 20.4 ft . . (d) 22.3ft

I have (c) 20.4 ft. .Is that correct? . Yes!

There are algebraic methods that could be used,
. . but with only 5 bounces, we can baby-talk our way through it.

$\text{First, it falls }4\text{ feet.\;\;[1st hit]}$

$\text{It bounces up }\tfrac{3}{4}(4) = 3\text{ feet and falls }3\text{ feet.\;\;[2nd hit]}$

$\text{It bounces up }\tfrac{3}{4}(3) = \tfrac{9}{4}\text{ feet and falls }\tfrac{9}{4}\text{ feet.\;\;[3rd hit]}$

$\text{It bounces up }\tfrac{3}{4}(\tfrac{9}{4}) = \tfrac{27}{16}\text{ feet and falls }\tfrac{27}{16}\text{ feet.\;\;[4th hit]}$

$\text{It bounces up }\tfrac{3}{4}(\tfrac{27}{16}) = \tfrac{81}{64}\text{ feet and falls }\tfrac{81}{64}\text{ feet.\;\;[5th hit]}$

$\text{The total distance is:}$

. . $D \:=\:4 + 2(3) + 2(\tfrac{9}{4}) + 2(\tfrac{27}{16}) + 2(\tfrac{81}{64}) \:=\:\frac{653}{32} \:=\:20.40625\text{ feet}$

Thanks