Question

3. 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? Round your answer to the nearest hundredth meter.
A) 208.02 m
B) 416.03 m
C) 0.52 m
D) 208.19 m

Answer 1

A) 208.02 meter

Explanation:

We are given,

The path of the rocket is modeled by
`y = -0.04x^2+8.3x+4.3`.

We have that the when the rocket lands, the value of y = 0.

Substituting y= 0, we get,

`y=-0.04x^2+8.3x+4.3\\\\0=-0.04x^2+8.3x+4.3\\\\0.04x^2-8.3x-4.3=0\\\\(x+0.517)(x-208.02)=0\\\\x=-0.517,\ x=208.02`

As 'x' represents the horizontal distance, it cannot be negative.

So, we have x= 208.02 meter.

That is, the rocket will land after 208.02 meters.

Thus, option A is correct.

Answer 2

When it hit the ground
y = 0
y = –0.04x^2 + 8.3x + 4.3
0.04x^2 - 8.3x - 4.3 = 0
x = [-(-8.3) +/- √{(-8.3)^2 - 4 x 0.04 x (-4.3)}]/[2 x 0.04]
x = [8.3 +/- 8.34]/[0.08]
x = 208.02 m
A is right option
hope this helps