Question

A ball is dropped from a height of 5 feet and bounces. Suppose that each bounce is 5/8 of the height of the bounce before. Thus, after the ball hits the floor for the first time, it rises to a height of 5 ({5\over 8}) = 3.125 feet, etc. (Assume g = 32 \hbox{ft/s}^2 and no air resistance.)

A. Find an expression for the height, in feet, to which the ball rises after it hits the floor for the nth time:
h_n =

B. Find an expression for the total vertical distance the ball has traveled, in feet, when it hits the floor for the first, second, third and fourth times:
first time: D =
second ti

Answer 1

the heights form a geometric series

Answer 2

\[5,5\left(5\over8\right),5\left(5\over8\right)^2,5\left(5\over8\right)^3,\ldots\\
h_n=5\left(5\over8\right)^n
\]

Answer 3

total vertical distance = sum fof the series

Answer 4

I got the first height being 5 feet but then I'm doing something wrong on the 2nd 3rd and 4th

Answer 5

each heach is
\(\displaystyle \left(5\over8\right)\)
of the previous height.

Answer 6

so the first is 5 and the second should be\[5\left(\begin{matrix}5 \\ 8\end{matrix}\right)\]

Answer 7

that doesn't work, so does the gravity have something to do with it?

Answer 8

??? did you pout that as a combination?

Answer 9

as in\[5+5\left( \frac{ 5 }{ 8 } \right)\]?

Answer 10

that doesn't work either

Answer 11

@electrokid are you still there?

Answer 12

second time, it'd be twice..
it has to go up and then come down the same distance