1st June, 2010
The puzzle is from famous puzzlest Sam Lyod.
A ball is dropped from building of 100 m height. The ball bounces 10 % of previous height due to Earth reaction. Question is how long the ball traveled before ceasing its motion?
If you get the answer as [h + h/10 + h/100 + h/1000 + so on], you have a lacuna in the problem understanding. Because except the first instance, the ball travels twice the distance of drop height. The solution fall into G. P. of infinite series. The answer is 122.22 m. Try it.
Here is a hint,
S(n) = h + h 2 + h 3 + h 4 + …
h x S(n) = h 2 + h 3 + h 4 + …
After simple subtraction we get the following simplification for S(n)
S(n) [1 – h] = h ==> S(n) = h/(1-h)
3 comments:
Let the height be 'H' m. The ball travels on total - H + 2H/10 + 2H/100 + 2H/1000 + ... meters
We can use infinite GP formula and H = 100 in the above series.
If we use the above formula the Answere would be 122.22 mts.
If we use the above formula the answere would be 122.22 meters.
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