Question

A ball is dropped from a height of 30 feet. The ball bounces. After each bounce, the maximum height of the ball is 80% of the previous height. Write an nth term formula to model the situation and approximate the maximum height of the ball after 6 bounces.

Answer 1

das is sum of geometric sequence
first is 30
imagine it
ball droppes 30 feet
bounces up 80% of that
then it has to bounce down again
then up
down
therfor, we summ up from 30 to 6thbounce, times 2 and minus 30 since the first bounce didn't start from ground (will include diagram)


so therefor
2(sum)-30 is the answer

the sum of a geometric sequence is
$S_{n}= \frac{a_{1}(1-r^{n})}{1-r}$
where Sn is the sum to the nth bounce
a1=first term
r=common ratio
n=which term
Sn=S6 (6th bonce)
a1=30
r=80% or 0.8

sub

$S_{6}= \frac{30(1-0.8^{6})}{1-0.8}$
$S_{6}= 110.678$

we then double then minus 30
2*110.678=221.357
minus 30
221.357-30=191.357 feet


you should use
dropped from x feet and max height is m percent of previous heigh, find total distance after n bounces (convert m% to decimal)
answer=$2[ \frac{x(1-m^{n}}{1-m} ]-x$





anyway, distance is 191.357 feet