To find where the rocket will land, we need to find when y=0 (the rocket will be at ground level). So we substitute y=0 into the equation and solve for x:
0 = -0.04x^2 + 8.3x + 4.3
Multiplying by -25 to simplify:
x^2 - 207.5x - 86 = 0
Using the quadratic formula:
x = (207.5 ± sqrt(207.5^2 - 4(1)(-86))) / 2(1)
x = (207.5 ± sqrt(42906.25)) / 2
The two solutions are x = 208.02 and x = -0.52. We can discard the negative solution because we are looking for distance, not direction. Therefore, the rocket will land 208.02 meters horizontally from its starting point, so the answer is A.
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
3 answers
are there any exponents in the model rocket's equation?
Yes, the equation for the path of the rocket involves an exponent: y = -0.04x^2 + 8.3x + 4.3. The term -0.04x^2 represents a quadratic function, where x is the independent variable and x^2 is the exponent.