 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  The Maypole Problem 2 (Posted on 2006-06-12) Continuing on the original problem here, posted by Jer.

During a gale a maypole was broken in such a manner that it struck the ground at a distance twenty feet from the base of the pole.

It was repaired without any gain or loss in its length. Later it broke a second time at a point below the earlier by an amount exactly equal to the difference in length of the two sections of the first break.

How far did it strike this time from the base?

 See The Solution Submitted by Salil Rating: 2.5000 (2 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Puzzle Solution Comment 3 of 3 | Since the length of the top section (l) and the length of the bottom (b) section during the first break satisfyl^2 - b^2 = 400, it follows that:
l>w, so that the respective lengths of the top and bottom section
during the first break are (a+b) and (a-b) (say).

So, (a+b)^2 - (a-b)^2 = 400, giving:
ab = 100

By the problem, the respective lengths of the top and bottom section during the second break are (a+b)+2b and (a-b)-2b

Let the distance that the maypole will strike this time
from the base  be S feet.

Then:
S^2
= ((a+b)+2b)^2 - ((a-b)-2b)^2
= (2a)*(6b)
=  12ab
= 12*100
= 1200, giving:
S = v(1200) = 34.641

Thus, the distance that the maypole will strike this time
from the base is 34.641 feet ( approx.).

Edited on May 11, 2007, 3:09 pm
 Posted by K Sengupta on 2007-05-11 15:09:05 Please log in:

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