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=-0.02x^2+2.3x+6, 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?

Answer 1

@dan815

Answer 2

@lunaone13 @Luigi0210

Answer 3

@133

Answer 4

Don't know so sorry!!!

Answer 5

the rocket landing point is at y = 0(ground level), so you'll have 2 possible Xs. Since the distance is a positive number, you'll take the positive x:
\[x _{1, 2} = \frac{ -2.3 \pm \sqrt{2.3^2 - 4*6*(-0.02)} }{ 2*(-0.02) } = \frac{ -2.3 \pm \sqrt{5.77} }{ 0.04 }\]

Answer 6

choices...
57.50m
115.00m
117.55m
235.10m

Answer 7

-0.04*

Answer 8

So would it be 117.55?

Answer 9

yep

Answer 10

WOO