A ball is dropped from a height of 12 ft. Each time it bounces....?

Question

it rises to a height of 60% of the distance it fell. How far does it travel by the end of its 5th bounce?

How do you solve this?

Answer 1

D = d0 + d sum(%b^N) for N = 1 to 5 bounces; where d = 12 ft. the initial height and %b = .6 and d0 = (12 or 24) depending on if raising the ball to height d = 12 is included in the total distance.

You can show this by:

N = 0 the starting point with d0 = (12 or 24 ft.), depending on if you count raising the ball to 12 ft. as well as dropping it to start the bouncing.
N = 1 the first bounce; so (.6)^1 = 60%; so the distance it travels here (d1) is d (%b) up AND also d (%b) down to start the next bounce. Thus, d1 = 2 d (%b^1)
N = 2 the second bounce; so (.6)*(.6) or 60% of the 60% first bounce; so the distance d2 = 2 d (%b)^2
N = 3 the third bounce distance would be similar; so d3 = 2 d (%b^3) and so on up to N = 5.

In which case, D = d0 + d1 + d2 + d3...+ d5 = d0 + 2 *12(.6 + .6^2 + .6^3 + .6^4 + .6^5) = d0 + 2 * 12 sum(.6^N) for N = 1 to 5; If you count raising the ball to the initial height, then d0 = 24 = 2 d; otherwise d0 = 12 ft., the distance of the first drop.

You can do the arithmetic.

Answer 2

the great is: h + h * 2/3 + (h * 2/3) * 2/3) + (h * 2/3) * 2/3 * 2 /3 + ... or sum(h * 2/3^n) this referred to as a geometrical progression: the place r < a million sum(h * r^n) = h / ( a million -r) = 12 / (a million - 2 / 3) = 36 I checked with a spreadsheet and interestingly good.

Answer 3

12X.6=7.2, 7.2X.6=4.32, 4.32X.6=2.592, 2.592X.6=1.5552, 1.5552X.6=0.93312, Then take 0.93312 and multiply it by the number of inches in a foot. So: 0.93312X12=11.19744 inches.

Answer 4

Just add them up, down 12, up 7.2, down 7.2, etc.