To find the parabolic equation that describes the height of the golf ball as it moves away from Lorna, we can model the height of the ball as a quadratic function of the horizontal distance from Lorna.
Since Lorna hits the ball vertically, we can assume that the vertex of the parabola is at the highest point of the ball's trajectory. The highest point the ball reaches, 30 ft, will be the vertex of the parabola. The parabola opens downward since the ball rises to a peak and then falls.
We can denote the horizontal distance from Lorna as \(x\) and the height of the ball as \(h(x)\). The vertex form of the quadratic function is given by:
\[ h(x) = a(x - h)^2 + k \]
where \((h, k)\) is the vertex of the parabola. In this case, the vertex is at \( (30, 30) \), indicating a horizontal value of \(30\) feet (halfway of the total 60 ft distance) and a height of \(30\) feet. We need to determine the value of \(a\).
To find \(a\), we can also use another point where the height of the ball is 0. We know that when the distance from Lorna is \(0\) ft (where she hits the ball), the height is also \(0\) ft:
\[ h(0) = 0 \rightarrow 0 = a(0 - 30)^2 + 30 \]
Plugging in \(0 = a(900) + 30\):
\[ 0 = 900a + 30 \] \[ 900a = -30 \] \[ a = -\frac{1}{30} \]
Now we can write the equation:
\[ h(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
Next, we will also use the other point where the height reaches \(0\) at a distance of \(60\) feet:
\[ h(60) = 0 \rightarrow 0 = -\frac{1}{30}(60 - 30)^2 + 30 \] \[ 0 = -\frac{1}{30}(30)^2 + 30 \] \[ 0 = -\frac{900}{30} + 30 \] \[ 0 = -30 + 30 \]
Thus, it is consistent.
Now, let’s find the height of the ball when \(x = 15\) feet:
\[ h(15) = -\frac{1}{30}(15 - 30)^2 + 30 \] \[ = -\frac{1}{30}(-15)^2 + 30 \] \[ = -\frac{1}{30}(225) + 30 \] \[ = -\frac{225}{30} + 30 \] \[ = -7.5 + 30 \] \[ = 22.5 \]
So the height of the ball when it is \(15\) ft away from Lorna is 22.5 ft.
Summary
- The parabolic equation is: \[ h(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
- The height of the ball when it is \(15\) ft away from Lorna is \(22.5\) ft.