Question

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

Answer 1

Option b. 215.56 is the right answer.

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
`x=\frac{-8.6\pm \sqrt{(8.6)^(2)-4(-.04)(4.8)}}{2* (-.04)}`

`=\frac{-8.6\pm \sqrt{8.6^(2)+4* .04* 4.8}}{2(-.04)}`

`=(-8.6\pm √(73.96)+.768)/((-.08))`

`=(-8.6\pm 8.64)/((-.08))`

`x=(-8.6-8.64)/(-.08)=215.5 meters`