A rubber ball dropped from a height of 40 cm rebounds on each bounce to a height which is 2/5 of its height on the previous bounce. How far will it travel before coming to rest?

Answer 1

This problem can be modeled with a sum. It is an infinite sum with smaller and smaller distances. Eventually 2/5 of the height will be so small the distance will be negligible.

It falls 40 cm on the drop. It would then rise 16 cm on the bounce (2/5 of 40) and another 16 cm back down. Then 32/5 cm up and 32/5 cm down... the pattern is as such:

40, 16, 16, 16*(2/5), 16*(2/5), 16*(2/5)^2, 16*(2/5)^2, .....

[*Note* there are two of every number except 40 because the ball rises and falls each time except on the initial drop]

Therefore the full sum distance is D = 40 + 2*Σ[16*(2/5)^n-1)] where n goes from 1 to infinity.

The 40 is for the initial drop. The multiplied factor of 2 is because there are two of each number aside from 40. Each time the 16 is successively multiplied by another 2/5.

Make sure you test this formula for yourself and see that is works.

The formula is a clear geometric sum. The infinite geometric sum formula is S = a*(1-r) where a is the first term and r is the successively multiplied factor. In this case r = 2/5 and a is 16. Calculate the sum and then complete the calculation using the formula I mentioned above: D = 40 + 2*S.

Note: S = Σ[16*(2/5)^n-1)] = a*(1-r) when summed to infinity

Answer 2

We have such a sum S = 40 + 2∑_n=1∞40(2/5)ⁿ= 40 + 2(16/(1 - 2/5)) = 40 + 32·5/3 = 40 + 160/3 = 280/3 cm

We use formula for sum of infinite geometric progression: S = a₁/(1 - r)

Answer 3

The first drop is 40 cm. Then, each bounce goes 2/5 as high up as the previous one went down and the same amount down. Note, then, that while 40 cm appears once in the sum, all others appear twice.

Thus, we have 2 * 40/(1 - 2/5) - 40 = 80 * 5/3 - 40 = 400/3 - 120/3 = 280/3 = 93 1/3 cm