A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation
y = –0.04x^2 + 8.3x + 4.3 , where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land? Round your answer to the nearest hundredth meter.
A. 208.02 m
B. 416.03 m
C. 0.52 m
D. 208.19 m
To find where the rocket lands, we need to find the x value when y = 0 (since the rocket will be at ground level). So, we set y = 0 and solve for x:
0 = –0.04x^2 + 8.3x + 4.3
Using the quadratic formula, we get:
x = (-8.3 ± sqrt(8.3^2 - 4(-0.04)(4.3))) / (2(-0.04))
x ≈ 208.19 or x ≈ 0.52
We can discard the x ≈ 0.52 solution since it corresponds to a point on the roof where the rocket was launched. Therefore, the rocket will land approximately 208.19 meters horizontally from its starting point. Rounded to the nearest hundredth, the answer is A. 208.02 m.