Calculus
posted by Sandy on .
A ball is dropped from a height of 10 feet and bounces. Each bounce is ¨ú of the height of the bounce before. Thus, after the ball hits the floor for the first time, the ball rises to a height of 10(¨ú ) = 7.5 feet, and after it hits the second floor for the second time, it rises to a height of 7.5(¨ú ) = 10(¨ú )©÷ = 5.625 feet. (Assume g= 32 ft/sec ©÷ and that there is no air resistance.)
(a) Find an expression for the height to which the ball rises after it hits the floor for the nth time.
(b) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the first, second, third, and fourth times.
(c) Find an expression for the total vertical distance the ball has traveled when it hits the floor for the nth time. Express your answer in closedform.
For part (a) I got ar^n
For part (b) I got
1st bounce = a + ar
2nd bounce = a + ar + ar + ar©÷
3rd bounce = a + ar+ ar + ar©÷ + ar©ø
4th bounce = a + ar +ar + ar©÷ + ar©ø + ar©ø + ar©ù
For part (c) I do not kow what to do any help would be greatly appreciated.

a) h = 10(.75)^n
that is basically what you had
b) d = 10 + 10(.75)^1 + 10 (.75)^2 + 10(.75)^3 .... In other words I sort of agree with you
c) This is a geometric series like a
compound interest problem
in general
g + gr + gr^2 + .... gr^(n1)
Sn = [ g (1r^n ]/ (1r)
here g = 10 and r = .75
so
Sn = [ 10 (1  .75^n) /.25 
I disagree with part b)
at the first bounce the ball has traveled 10 ft.
at the second bounce, it went up 7.5 and then down 7.5, so
distance after two bounces = 10 + 2(10)(.75)^1
after 3 bounces it went 10 + 2(10)(.75) + 2(10)(.75)^2
so after the first bounce, you have to double the distance, since it goes up and then the same distance down again.
so the series
= 10 + 2(10)(.75) + 2(10)(.75)^2 + 2(10)(.75)^3 + ....
so
Sn = 10 + 20(.75)[1  .75^(n1)]/(1  .75)
so I see an infinite geometric series starting with the second term and that sum is
10 + (20)(.75)/(1  .75)
= 10 + 15/.25
= 70 feet