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A rubber ball has the property that, on any given bounce, it returns to one-third of the height from which it just fell. Suppose the ball is dropped from 108 feet. How far has the ball traveled the fourth time it hits the ground?

2006-12-15 09:09:04 · 3 answers · asked by TheRenegade 1 in Science & Mathematics Physics

3 answers

108 + 108/3*2 + 108/9*2+ 108/27*2

2006-12-15 09:15:32 · answer #1 · answered by Snowflake 7 · 0 1

You mean as in:
108 feet (boing #1)
+ (1/3)*108 upwards + (1/3)*108 down (boing #2)
+ (1/3)*(1/3)*108 upwards + down (boing #3)
+ up + down (boing #4)...

2006-12-15 17:14:16 · answer #2 · answered by Raymond 7 · 1 1

assuming that it is only travelling in the vertical direction (z axis) then your equation should look like this:

D = X + (0.33)X + (0.33)(0.33)X + (0.33)(0.33)(0.33)X + (0.33)(0.33)(0.33)(0.33)X

where X is the height from which you drop the ball. The general equation would look like this:

D = X + (the sum of all values from 1 to n of general equation) X(0.33)^N

where n = the number of bounces you're looking for and X equals the height dropped

2006-12-15 17:14:59 · answer #3 · answered by promethius9594 6 · 0 1

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