geometric progression/sequence???

Question

a rubber ball was dropped from a height of 42m and each time it strikes the ground, it rebounds to a height of 2/3 of the distance from which it fell. find the total distance traveled by the ball before it comes to rest...step by step please...thanks

a rubber ball was dropped from a height of 42m and each time it strikes the ground, it rebounds to a height of 2/3 of the distance from which it fell. find the total distance traveled by the ball before it comes to rest...step by step please...thanks

the answer is 210 m

Answer 1

The sum of a geometric sequence is

Sn = a1( 1- r^n)/( 1- r)
if |r| < 1 then for S = a1/(1-r)

so for your problem a1 = 42 and r = 2/3

so total distance = 42/(1-2/3) = 126

Answer 2

The ball fell 42m, so there is a start.

It rebounded 2/3 of that distance, or 28m...of course it then fell back to the ground, so double that distance for 56m.

So...each time the ball travells 2*(2/3)*the previous height.

42 + 2*(2/3)*42 + 2*(2/3)^2*42 + ... =

42 + 56*[1 + (2/3) + (2/3)^2 + ...]

The part in brackets is a geometric sum, for which you may use the formula 1/(1-r).

Answer 3

42m + 28m + 28m + 18 2/3m+18 2/3m +12 4/9+12/4/9 +...
= 42 + 4/3(42) +8/9(42) + 16/27(42) + .. 4n/3^n(42) + ...
= 42 +42(4/3 +8/9 +16/27 +... 4n/3^n +...)
4n/3^n = 0, so the series converges.
lim n --> infinity
Summing the 1st10 terms will give you an answer accurate to 3 decimal places. You can do this easily on a spreadsheet.

Answer 4

So we are just summing a bunch of distances:

42+42*(2/3)+42*(2/3)*(2/3)+42*(2/3)*(2/3)*(2/3)+........
=42+42*(2/3)+42*(2/3)^2+42*(2/3)^3........

We know from the formula for the sum of a convergent series that the above sum is equal to the following:

42/(1-(2/3))=3*42=126

Answer 5

42 + 2 * [28/(1 - 1/3)] = 42 + 2 * [42] = 126 meters

[Remember, it goes up and down the height, hence the 2x.]