Asked by Beth
When a hammer thrower spins, the hammer generates a circle with a radius of 5 feet. When thrown, the hammer hits a screen that is 50 feet from the center of the throwing area. Let coordinate axes be introduced. If the hammer is released at (-4, -3) and travels in the tangent direction, where will it hit the screen?
I graphed the situation and found the answer to be ((31.5,-50). I want to learn how to figure it out mathematically so I can then figure out where the hammer should be released in order for the hammer to hit the screen at (0, -50).
I graphed the situation and found the answer to be ((31.5,-50). I want to learn how to figure it out mathematically so I can then figure out where the hammer should be released in order for the hammer to hit the screen at (0, -50).
Answers
Answered by
Steve
Man, there's a lot of missing information here. That leads me to conclude that the hammer spins in a counterclockwise direction, then flies in a straight, horizontal line from the circle to the screen. Apparently the screen is a circle of radius 50.
Now, your solution of (31.5,-50) is not 50 units from (0,0).
So, please elucidate a bit more on just what's going on here.
If I have it right, you want to release the hammer at (h,k) so that the tangent line hits the screen at (0,-50)
At (h,k), the tangent line has slope -h/k, so the line is
y-k = -h/k (x-h)
But, we know that h^2+k^2 = 25, so
y-√(25-h^2) = -h/√(25-h^2) (x-h)
and that line must pass through (0,-50).
-50-√(25-h^2) = -h/√(25-h^2)(-h)
h = -4.97
Hmm, that sends the hammer on a path that is nearly vertical, so you'd better check my math.
Now, your solution of (31.5,-50) is not 50 units from (0,0).
So, please elucidate a bit more on just what's going on here.
If I have it right, you want to release the hammer at (h,k) so that the tangent line hits the screen at (0,-50)
At (h,k), the tangent line has slope -h/k, so the line is
y-k = -h/k (x-h)
But, we know that h^2+k^2 = 25, so
y-√(25-h^2) = -h/√(25-h^2) (x-h)
and that line must pass through (0,-50).
-50-√(25-h^2) = -h/√(25-h^2)(-h)
h = -4.97
Hmm, that sends the hammer on a path that is nearly vertical, so you'd better check my math.
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