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 + 8.6x + 4.8, 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. 0.56 m
b. 215.56 m
c. 431.11 m
d. 215.74 m

Question

contestada

Respuesta :

eudora eudora · Answer 1 · 2019-02-03T23:01:58+01:00

Answer:

Option b. 215.56 is the right answer.

Step-by-step explanation:

Path of the rocket can be modeled by the equation y = -0.04x²+8.6x+4.8

Here y represents the vertical height of the rocket and x represents the horizontal distance in meters.

We have to calculate the horizontal distance covered by the rocket.

Now we can say when rocket is landing y = 0 or vertical distance of the rocket will be zero.

Then equation will be

-.04x²+8.6x+4.8 = 0

Then [tex]x=\frac{-8.6\pm \sqrt{(8.6)^{2}-4(-.04)(4.8)}}{2\times (-.04)}[/tex]

[tex]=\frac{-8.6\pm \sqrt{8.6^{2}+4\times .04\times 4.8}}{2(-.04)}[/tex]

[tex]=\frac{-8.6\pm \sqrt{73.96}+.768}{(-.08)}[/tex]

[tex]=\frac{-8.6\pm 8.64}{(-.08)}[/tex]

[tex]x=\frac{-8.6-8.64}{-.08}=215.5 meters[/tex]