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Jack's Ladder (Posted on 2004-07-27) Difficulty: 2 of 5
Even though it was now middle of winter, Jack hauled out his ladder and placed his ladder against the side of the house and began to climb. He had some bad luck when he reached the half-way point. The ground was a frozen sheet of ice and the base of his ladder slipped out and the top slid down the side of the house. Jack, clinging to the center rung, wound up moving from Point A on the side of his house to Point B on the ground.

Describe the path Jack traveled.

See The Solution Submitted by Brian Smith    
Rating: 2.6667 (3 votes)

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re(3): Good explanation | Comment 19 of 22 |
(In reply to re(2): Good explanation by Jim)

OK, assume jack has a 10' ladder (just to make it easy).

If the ladder is magically stuck against the wall (maybe frozen), the top of the ladder (y axis coordinate) is 10, and the bottom of the ladder (x axis coordinate) is 0. jack is in the middle of the ladder, so his coordinates are (0,5).

As the ladder slips 1 foot down the wall (y=9), the bottom of the ladder has to come out 4.36 feet from the wall (x=4.36). Jack's coordinates are (2.18,4.5) as he is still in the middle of the ladder.

As y keeps dropping 1 foot at a time, you get the following....

Y=8, X=6, JACK = (3,4)

Y=7, X=7.14, JACK = (3.57,3.5)

Y=6, X=8, JACK = (4,3)

Y=5, X=8.66, JACK = (4.33,2.5)

Y=4, X=9.16, JACK = (4.58,2)

Y=3, X=9.54, JACK = (4.77,1.5)

Y=2, X=9.8, JACK = (4.9,1)

Y=1, X=9.95, JACK = (4.97,1/2)

Y=0, X=10, JACK = (5,0)

If you plot these points, you definitely get an arc that LOOKS like a 1/4 circle. But if anyone knows the formula for a circle, radius 10, offset 0, please post and run these values.

  Posted by Jim on 2004-08-10 09:17:20
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