Question

help please

Answer 1

A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y= -0.8^2+12x+25.8 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? Round your answer to the nearest hundredth.

Answer 2

answer choices:

25.80 m

37.00 m

17.24 m

16.91 m

Answer 3

Note that this problem does not focus on TIME (t) at all, but only on vertical and horiz. distances.

y represents the distance of the model rocket above the ground. When the rocket hits the ground, y will obviously equal zero (0).

Set the given equation = to 0 and solve for x. This x is the horiz. distance that you want.

Answer 4

wait, so I'm a little confused. I don't get this at all ((sorry!!)) How would I set the equation up to solve it? @mathmale

Answer 5

Take the given equation and set it equal to zero. It'll become a "quadratic equation." How have you solved quadratic equations in the past? Apply your preferred method here. Find the value or values of x that make the most sense.

Answer 6

So the given equation is y= -0.8^2+12x+25.8? I think I've solved them before but not recently so that is why I am having all this trouble.

Answer 7

@mathmale

Answer 8

Which methods have you used in the past to solve quadratic equations?

Answer 9

graphing
factoring
quadratic formula
identification of vertex
completing the square
and so on

Answer 10

Set y= -0.8^2+12x+25.8 equal to zero and begin solving for x using the method of your choice.

Answer 11

I tried doing the quadratic formula and I got x=−2.096667 after doing -0.8^2+12x+25.8=0 @mathmale

Answer 12

Does that sound about right? @mathmale

Answer 13

Does a negative result strike you as being reasonable? Remember, x represents a horizontal distance. Weren't there TWO solutions?

Answer 14

What were your coefficients, a, b and c, to plug into the quadratic formula?

Answer 15

I don't even know, I doubt I even did it right because I have no idea how to do any of these in the first place. I haven't done quadratic equations for awhile now. I barely even know half the words you are using that is why I need help or I need help breaking it down. I can't really go off anything else.

Answer 16

-0.8^2+12x+25.8=0 yields the coefficients -0.8, 12 and 25.8. If you'd prefer not to deal with fractions, then multiply all of these by 10.

Answer 17

supposing that you're given the "quadratic equation" \[y=ax^2+bx+c=0,\]the quadratic formula enables you to calculate the x values that make this equation true:

Answer 18

\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 19

Have you seen this before?

Answer 20

Whether or not you have, if y ou are given the values of a, b and c, would you know what to do next, to find the "roots" or "solutions" \[x _{1}~and~x _{2}?\]

Answer 21

the quadratic formula, given earlier, is \[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 22

In y our equation, a=-0.8, b=12, and c=25.8. Replace a, b and c in the formula with these a, b and c values and calculate x. Can you do this?

Answer 23

Yes I get it now I think

Answer 24

\[x=\frac{ -12 \pm \sqrt{12^2-4(-0.8)(25.8)} }{ 2(-0.8) }\]

Answer 25

Evaluate x (two different values). If there's a negative value, throw it out and keep the positive one. That x value will represent the horizontal distance you wanted.

Answer 26

You must present your answer in the following format: dd.xx (express your answer accurate to two decimal places).