Question

A ball is dropped from a height of 41m. Each time it bounces it reaches 85% of it's previous height. When it hits the ground for the 26th time it has traveled Answerm.
(Give answer to 2 decimal places)

Answer 1

Hmm lets see, does this want total distance traveled from the beginning? Or just how high the ball is in difference to the beginning.

I believe the prior,

So the common ratio is going to be .85

a1 (the beginning height) will be 41 m

So the sum of the total distance traveled will be solved using

\[\large S_{26} = \frac{a_1(1 - r^n)}{1 - r}\]

so plugging in everything you have

\[\large S_{26} = \frac{41(1 - (.85)^{26})}{1 - .85}\]
\[\large S_{26} = \frac{41(1 - (0.01461813837))}{.15}\]
\[\large S_{26} = \frac{41(0.98538186163)}{.15}\]
\[\large S_{26} = \frac{40.4006563268}{.15}\]
\[\large \S_{26} = 269.34\]

This will be the TOTAL distance that ball has travelled from the drop, to the 26th bounce

Answer 2

The distance traveled between two successive bounces will be twice the height reached. Therefore the sequence in terms of distance traveled is of the form:
\[a,\ 2ar,\ 2ar ^{2},\ 2ar ^{3},\ 2ar ^{4},.........,\ 2ar ^{25}\]
The total distance traveled is given by:
\[\large \frac{2\times41(1-0.85^{26})}{1-0.85}-\frac{41}{1}\]
Note that 41 is subtracted because the distance of 41 m is traveled only once.

Answer 3

thanks guys! :D

Answer 4

So I calculate the total distance traveled to be 497.7 m.

Answer 5

what why are the 2 answers different o.o

Answer 6

Ask @johnweldon1993 to list the terms of his sequence in distances traveled (as I have done). Then compare the two sequences.