SOLUTION: A Special rubber ball is dropped from a height of 128 meters . Each time the ball bounces to half its previous height. After 1 bounce the ball reaches a height of 64 meters.
a.
Algebra ->
Customizable Word Problem Solvers
-> Misc
-> SOLUTION: A Special rubber ball is dropped from a height of 128 meters . Each time the ball bounces to half its previous height. After 1 bounce the ball reaches a height of 64 meters.
a.
Log On
Question 195215: A Special rubber ball is dropped from a height of 128 meters . Each time the ball bounces to half its previous height. After 1 bounce the ball reaches a height of 64 meters.
a. How high did the ball reach after the 7th bounce.
b.How high does the ball get after the 9th bounce Answer by windsolace(8) (Show Source):
You can put this solution on YOUR website! The series identified is a geometric progression with common ratio of half.
Formula for geometric progression:
Nth term of series = ar^(n-1), where a = first term value, r = common ratio
a) 7th term of the series = 64(1/2)^(7-1) = 64(1/2)^6
= 64(1/64)
= 1
The ball reached a height of 1m after the 7th bounce
b) 9th term of series = 64(1/2)^8
= 64/256
= 0.250
The ball reached a height of 0.25m (25cm) after the 9th bounce.
(