To solve this problem, we can represent the height of the ball as a quadratic function. Given the information provided:
- The ball reaches a maximum height of 30 feet (the vertex of the parabola).
- The ball lands 60 feet away from Lorraine (the x-intercepts of the parabola).
- The ball starts at 0 feet away from Lorraine (the starting point).
1. Determine the vertex and axis of symmetry
The vertex of the parabola, which represents the maximum height of the ball, will have the x-coordinate at half of the total distance:
- The x-coordinate of the vertex is \( \frac{60}{2} = 30 \).
- Hence, the vertex is at \( (30, 30) \).
2. Identify points for the quadratic equation
The two points where the ball lands are (0, 0) and (60, 0) since it lands 60 feet away. The vertex is (30, 30).
Using the vertex form of the quadratic equation: \[ f(x) = a(x - h)^2 + k \] where \( (h, k) \) is the vertex.
Plugging in the vertex (30, 30): \[ f(x) = a(x - 30)^2 + 30 \]
3. Determine the value of \( a \)
To find \( a \), we can use one of the other points, let’s use the point where the ball lands, \( (0, 0) \): \[ 0 = a(0 - 30)^2 + 30 \] Solving for \( a \): \[ 0 = a(900) + 30 \] \[ a(900) = -30 \] \[ a = -\frac{30}{900} = -\frac{1}{30} \]
4. Final equation
Combining this value back into the vertex form gives us: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
Question 10: Write the equation
The equation for this scenario is: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
Height of the ball when it is 15 feet away from Lorraine
To find the height of the ball when it is 15 feet away (x = 15), we substitute \( x = 15 \) into our equation: \[ f(15) = -\frac{1}{30}(15 - 30)^2 + 30 \] Calculating this: \[ f(15) = -\frac{1}{30}(-15)^2 + 30 = -\frac{1}{30}(225) + 30 \] \[ f(15) = -\frac{225}{30} + 30 = -7.5 + 30 = 22.5 \]
Summary
- The equation is: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
- The height of the golf ball when it is 15 feet away from Lorraine is 22.5 feet.
1 answer