To solve this problem, we need to derive the equation of the parabola based on the given information.
Given Information:
- The ball is hit from point \( (0, 0) \), which represents Lorraine's position.
- The vertex of the parabola is at \( (30, 0) \) (the highest point).
- The ball lands 60 feet away, so the x-intercepts are at \( (0, 0) \) and \( (60, 0) \).
- The height at the vertex \( (30, 30) \) represents the maximum height of the ball.
Step 1: Find the Vertex
The vertex of the parabola in standard form is at \( (h, k) \), where \( (h, k) = (30, 30) \).
Step 2: Determine the Leading Coefficient \( a \)
The general equation of a parabola that opens downwards can be expressed as: \[ f(x) = a(x - h)^2 + k \] Substituting for the vertex \( (30, 30) \): \[ f(x) = a(x - 30)^2 + 30 \]
Step 3: Use the x-intercepts to find \( a \)
Since the ball lands at \( (0, 0) \), we can use this point to find the value of \( a \): \[ 0 = a(0 - 30)^2 + 30 \] \[ 0 = 900a + 30 \] \[ 900a = -30 \] \[ a = -\frac{30}{900} = -\frac{1}{30} \]
Step 4: Write the Equation
Now substituting \( a \) back into the equation: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
Question 10 Answers:
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The equation for this scenario is: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
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Now, to find the height of the ball when it is 15 feet away from Lorraine (\( x = 15 \)): \[ f(15) = -\frac{1}{30}(15 - 30)^2 + 30 \] \[ = -\frac{1}{30}(-15)^2 + 30 \] \[ = -\frac{1}{30}(225) + 30 \] \[ = -\frac{225}{30} + 30 \] \[ = -7.5 + 30 \] \[ = 22.5 \] Thus, the height of the ball when it is 15 feet away from Lorraine is approximately: \[ 22.5 \text{ feet in the air} \]
Summary of Results:
- The equation is: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
- The height of the ball when it is 15 feet away from Lorraine is: \[ 22.5 \text{ feet} \]
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