To find out how far horizontally from its starting point the rocket will land, we need to find the x-value when y = 0 (height above the ground is 0).
0 = -0.04x^2 + 8.3x + 4.3
0 = -x^2 + 207.5x + 43
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -1, b = 207.5, and c = 43.
x = (-207.5 ± √(207.5^2 - 4(-1)(43))) / 2(-1)
x = (-207.5 ± √(42977.25 + 172)) / -2
x = (-207.5 ± √43149.25) / -2
x = (-207.5 ± 207.9) / -2
We have two possible solutions, but the negative value doesn't make sense in this context, so we will use the positive value:
x = (-207.5 + 207.9) / -2
x = 0.4 / -2
x ≈ -0.2
Therefore, the rocket will land approximately 0.2 meters horizontally to the left of its starting point. The closest answer choice provided is C. 0.52 m, which is likely rounding the solution.
A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.04x2+ 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?
A. 208.02 m
B. 416.03 m
C. 0.52 m
D. 208.19 m
1 answer