Question

a model rocket is launched from a roof into a large field. the path of the rocket can be modeled by the equation -0.02x2+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 will the rocket land?

a) 57.50m
b) 115.0m
c) 117.55m
d) 235.10m
PLEASE ANSWER AND REPLY ASAP! THANK YOU GUYS A LOT!!

Answer 1

I have a feeling that some information is missing from the question.
It gives f(x)=y=-0.02x2+2.3x+6
and gives 4 possible values of x without supplying the value for y (or f(x)).]

Answer 2

how far \(\bf horizontally\) will the rocket land? what's that asking for?

Answer 3

ITS ASKING for how far horizontally it will land from the starting point

Answer 4

do you get it now?

Answer 5

It will depend on
1. the initial velocity
2. the initial angle with the horizontal
3. the height of firing point from the ground
We do not have any of the above information, nor any information related to the maximum height (y), so any horizontal distance is possible. Could you double check your question to see if something is missing, e.g. a diagram/figure, initial data previous to the question, general assumptions (if it is part of a long question), etc.

Answer 6

ok wait

Answer 7

a model rocket is launched from a roof into a large field. the path of the rocket can be modeled by the equation y= -0.02x2 + 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 8

Do you have a figure that shows how high the roof was?

Answer 9

notice the equation is a quadratic equation and thus a parabola
notice the leading term coefficient is negative, meaning is opening downwards
\(\bf {\color{brown}{ y}}=-0.02x^2+2.3x+6 \implies {\color{brown}{ 0}}=-0.02x^2+2.3x+6\)

solve for "x", or horizontal distance