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).

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.

To figure out where the hammer will hit the screen, we can use the concept of similar triangles and trigonometry.

First, let's set up the problem using the coordinate axes. The center of the throwing area is the origin (0,0) on the coordinate plane.

Based on the information given, we can determine that the radius of the circle generated by the hammer is 5 feet, which means the hammer will travel along a circular path with a radius of 5 feet.

The point where the hammer is released is (-4, -3). Let's call this point A.

The point where the hammer hits the screen is the point we want to find. Let's call this point B and assume it will be somewhere along the x-axis.

To find point B, we need to calculate the x-coordinate relative to the origin. Since the hammer is released from point A and moves in a tangent direction, we need to find the slope of the tangent line at point A.

The slope of the tangent line can be calculated using the derivative of the equation of the circle. The equation of a circle with radius 5 centered at the origin is x^2 + y^2 = 5^2.

Taking the derivative with respect to x, we get:
2x + 2y(dy/dx) = 0

Since we know that (x=-4, y=-3) is a point on the tangent line, we can substitute these values into the equation:
2(-4) + 2(-3)(dy/dx) = 0
-8 - 6(dy/dx) = 0
-6(dy/dx) = 8
dy/dx = -8/(-6)
dy/dx = 4/3

Now that we have the slope of the tangent line, we can find the equation of the tangent line passing through point A. Using the point-slope form, the equation becomes:
(y - (-3)) = (4/3)(x - (-4))
y + 3 = (4/3)(x + 4)

Next, we need to calculate the x-coordinate where the hammer will hit the screen. We know that the screen is located 50 feet from the center of the throwing area, which is the origin. Given that the y-coordinate of point B is -50 (on the screen), we can substitute these values into the equation we found:
-50 + 3 = (4/3)(x + 4)
-47 = (4/3)(x + 4)

To solve for x, we can multiply both sides of the equation by 3/4 to isolate x:
-47 * (3/4) = x + 4
-70.5 = x + 4
x = -70.5 - 4
x = -74.5

So, the x-coordinate where the hammer hits the screen is -74.5.

Therefore, the final answer is the point B, which is (-74.5, -50).