To create an equation representing the height of the golf ball in relation to its distance from Lorraine, we can use the vertex form of a quadratic equation. Since the ball reaches a maximum height of 30 feet at its highest point, we know that the vertex of the parabola is at the point (30, 30).
From the problem, we know that the ball lands 60 feet away from Lorraine, meaning the x-intercepts (the points where the ball touches the ground) are at x = 0 (when it is hit) and x = 60 (when it lands).
To create the function, we can factor it based on the roots: \[ h(x) = a(x)(x-60) \] Where \( h(x) \) is the height of the ball, \( x \) is the horizontal distance from Lorraine, and \( a \) is a scaling factor.
Given that the maximum height of the ball is 30 feet, we can find \( a \):
- The vertex (30, 30) implies the height at \( x = 30 \) is 30.
- Substitute into the equation: \( h(30) = a(30)(30-60) = 30 \).
Calculating: \[ h(30) = a(30)(-30) = -900a \] Setting this equal to 30 gives us: \[ -900a = 30 \implies a = -\frac{30}{900} = -\frac{1}{30} \]
Now we can write the complete equation: \[ h(x) = -\frac{1}{30}(x)(x-60). \]
To get the standard form of the equation: \[ h(x) = -\frac{1}{30}(x^2 - 60x) = -\frac{1}{30}x^2 + 2x. \]
Next, we need to evaluate the height of the golf ball when it is 15 feet away from Lorraine (i.e., \( x = 15 \)): \[ h(15) = -\frac{1}{30}(15^2) + 2(15) \] Calculating \( h(15) \): \[ = -\frac{1}{30}(225) + 30 = -\frac{225}{30} + 30 = -7.5 + 30 = 22.5 \text{ feet}. \]
Therefore, the height of the golf ball when it is 15 feet away from Lorraine is: \[ \text{Answer: } 22.5 \text{ feet.} \]