I have a wire that is hanging 40 feet above the ground.This wire has a circle that is 9 feet in radius around it. There is a forest to one side of the line. The forest starts 20 feet away (horizontal distance from where the wire hangs). I need to figure out how tall a tree needs to be to pass within the circle if it fell using any distance outside the 20 feet....
Is a tree tall enough to hit a target that is above the ground
0
$\begingroup$
geometry
-
0What do you mean "circle around it" ? – 2012-11-01
1 Answers
1
It's almost as simple as Pythagorean triangle equation:
- $ y $ - height of a wire
- $ x $ - distance of a tree
- $ r $ - radius of circle O around wire
- $ h $ - height of a tree
Shortest possible tree to pass within limits of circle O is the one which has radius of "falling circle" such that this "falling circle" (lets call it F) is exactly tangent to circle O. Then:
$ x^2 + y^2 = (h+r)^2 $
So:
$ h = \sqrt {x^2+y^2} - r $
So after using measurements from question:
$ h = \sqrt {40^2+20^2}-9 = \sqrt {2000} - 9 \approx 44.721 - 9 = 35.721 $
Edit
Maybe you are wondering "what if $x>20$? Is it possible that $h$ will be smaller?"
It can be seen that both $ \sqrt x$ and $x^2 + 1600$ are growing functions - so composition of those two is also always getting bigger. You can also look at this plot: