Question

The cup, on the 9th hole of a golf course, is located, dead center, in the middle of a circular green, which is 35 feet in radius. The ball follows a straight-line path, and exits the green at the right-most edge. Assume the ball travels 11 ft/sec. Introduce coordinates, so that the cup is the origin of an xy-coordinate system. Provide numerical answers below, with two decimal places of accuracy. Ball starts at (-40,-50). Suppose that L is a line, tangent to the boundary of the golf green, and parallel to the path of the ball. Let Q be the point where the line is tangent to the circle. Notice that there are two possible positions for Q. Find the possible x-coordinates of Q.

Answer 1

Answer:

± 27.33 ft

Step-by-step explanation:

For the given problem, we can estimate the initial and final coordinates of the line of the ball path as (-40,-50) and (0,0). Therefore, the slope is:

(-50-0)/(-40-0) = 50/40 = 1.25

Similarly, we can estimate the slope of a perpendicular line to the line of the ball path as: -1*(1/1.25) = -0.8.

Therefore, using (0,0) and the slope -0.8, the equation of the perpendicular line is: -0.8 = (y-0)/(x-0);

-0.8 = y/x

y = -0.8x

Furthermore, we are given the circle radius as 35 ft and we can use the distance formula to find the two points 35 ft far from the origin:

35^2 = x^2 + y^2

y = -0.8x

35^2 = x^2 + (-0.8x)^2

1225 = (x^2 + 0.64x^2)

1225 = 1.64x^2

x^2 = 1225/1.64 = 746.95

x = sqrt(746.95) = ± 27.33 ft

Answer 2

A circle is formed by all points on a plane equidistant from a given point

The possible x-coordinates of Q are -19.41 and +19.41

Reason:

Known parameter are;

Radius of the golf course, r = 35 feet

Point the ball exits the green = The right-most edge

Speed of the ball = 11 ft./sec

Location of the cup = The origin (0, 0)

Location the ball starts = (-40. -50)

Required:

To find the possible x-coordinates of Q, the point where the line L parallel to the path of the ball is tangent to the circle

Solution:

The coordinates of the rightmost edge of the golf course = (35, 0)

$Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$Slope \ of \ the \ path, \, m =\dfrac{-50-0}{-40-35}=\dfrac{-50}{-75} = \dfrac{2}{3}$

The equation of a circle is (x - h)² + (y - k)² = r²

The center of the circular green, (h, k) = (0, 0)

The equation of the given circle is (x - 0)² + (y - 0)² = x² + y² = 35²

$Slope \ of \ the \ radius \ at \ the \ tangent \ point, \ m_r = -\dfrac{1}{m}$

Equation of the radius at tangent point is therefore;

• $y = -\dfrac{3}{2} \cdot x$

Which gives the equation of the circle as follows;

• $x^2 + \left(-\dfrac{3}{2} \cdot x\right)^2 = 35^2$

$\dfrac{13 \cdot x^2}{4} = 35^2$

$x = \sqrt{ \dfrac{35^2 \times 4}{13} }= \pm \dfrac{70}{13} \cdot \sqrt{13} \approx \pm 19.41$

The possible x-coordinates of the point Q are -19.41 and +19.41

Learn more about the equation of a circle here:

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