To find the equation of the parabolic trajectory of the golf ball, we assume the shape of the path follows the form \( f(x) = ax^2 + bx + c \).
Given the information:
- The ball lands 60 feet away, so we can think of the endpoints as \( (0, 0) \) and \( (60, 0) \) (where \( y \) is the height).
- The maximum height of the ball is 30 feet, which occurs at the vertex of the parabola.
Since the highest point is at 30 feet, it will occur at the midpoint of the horizontal distance. The midpoint of \( 0 \) and \( 60 \) is \( 30 \). Thus, the vertex is at \( (30, 30) \).
The general form of the parabola can also be expressed using the vertex form as: \[ f(x) = a(x - h)^2 + k \] where \( (h, k) \) is the vertex. Here, \( h = 30 \) and \( k = 30 \), so we have: \[ f(x) = a(x - 30)^2 + 30 \]
Next, we can find the coefficient \( a \) using one of the points where the ball lands. We know that when \( x = 0 \) (the starting point), \( f(0) = 0 \): \[ 0 = a(0 - 30)^2 + 30 \] \[ 0 = a(900) + 30 \] \[ -30 = 900a \] \[ a = -\frac{1}{30} \]
Thus, the equation becomes: \[ f(x) = -\frac{1}{30}(x - 30)^2 + 30 \]
We can expand this equation to express it in standard form: \[ f(x) = -\frac{1}{30}(x^2 - 60x + 900) + 30 \] \[ f(x) = -\frac{1}{30}x^2 + 2x - 30 + 30 \] \[ f(x) = -\frac{1}{30}x^2 + 2x \]
So, the final equation of the trajectory of the golf ball is: \[ f(x) = -\frac{1}{30}x^2 + 2x \]
1 answer