A golf ball is hit so that it travels a horizontal distance of 500 feet and reaches a maximum height of 160 feet.a) Determine a quadratic equation that models the path of the golf ball, assuming it starts at the origin. (Round the coefficient of x2 to the nearest ten-thousandths place, and round the coefficient of x to the nearest thousandths place.)
Question
A golf ball is hit so that it travels a horizontal distance of 500 feet and reaches a maximum height of 160 feet.a) Determine a quadratic equation that models the path of the golf ball, assuming it starts at the origin. (Round the coefficient of x2 to the nearest ten-thousandths place, and round the coefficient of x to the nearest thousandths place.)
Solution
The path of the golf ball can be modeled by a quadratic equation of the form y = ax^2 + bx + c, where y is the height of the ball, x is the horizontal distance, and a, b, and c are constants.
Given that the ball starts at the origin, we know that c = 0.
We also know that the ball reaches a maximum height of 160 feet at half the horizontal distance, which is 250 feet. This gives us the vertex of the parabola, (h, k), where h = 250 and k = 160.
The standard form of a parabola is y = a(x - h)^2 + k. Substituting the values of h and k, we get y = a(x - 250)^2 + 160.
We also know that the ball lands at a horizontal distance of 500 feet, which gives us another point on the parabola, (500, 0). Substituting these values into the equation, we get 0 = a(500 - 250)^2 + 160.
Solving for a, we get a = -160 / (250)^2 = -0.00256 (rounded to the nearest ten-thousandths place).
Substituting the value of a back into the equation, we get y = -0.00256(x - 250)^2 + 160.
Expanding this equation, we get y = -0.00256x^2 + 1.28x + 160 (rounded to the nearest thousandths place).
So, the quadratic equation that models the path of the golf ball is y = -0.00256x^2 + 1.28x.
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