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+5.8x+4.9, 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?
Also not trying to be pushy but this is due tonight and no one in my family seems to know how to help me because they don't understand algebra, so can someone help me please?
This question is multiple choice:
a. 146.12 m
b. 291.68 m
c. 145.84 m
d. 0.84 m
Sorry I'm late! For anyone seeing this question, the answer is A - 208.02 m !
This question requires that you know how to work the quadratic equation formula.
I don't understand why you would be doing these questions without knowing that topic.
Horizontal distance = x
Vertical height = y
x is desired; therefore, y is neglected, meaning that you should set y = 0
Given: y = -0.02(x)^2 + 2.3(x) + 6
Setting y = 0
0 = -0.02(x)^2 + 2.3(x) + 6 <-----------> ax^2 + bx + constant = 0
Note: a = -0.02, b = 2.3, c = 6
Use the quadratic formula:
x = -(b) ± [(b)^2 - 4(a)(c)] ^(1/2) ... all over 2(a)
x = -(2.3) ± [(2.3)^2 - 4(-0.02)(6)]^(1/2)... all over 2(-0.02)
x = [-2.3 ± √5.77] / -0.04
You get one value of x when you add the square route of 5.77:
x = -2.5521 meters approx.
You get another value of x when you subtract the square route of 5.77:
x = 117.55 meters approx.
And we can neglect the negative value of x since we are talking about real-world progression.
so... x = 117.55 meters
Check your work:
0 ?=? -0.02(117.55)^2 + 2.3(1117.55) + 6
0 = 0.00495
Set y = 0 because that is when the rocket hits the ground. The x will be the horizontal difference.
Solve by using the quadratic equation. a= -0.04 b=5.8 c =4.9
x = -b +/- sq rt of(b^2 - 4ac) all of this is divided by 2a.
You will want the positive answer.
You want to know what x, defined as the horizontal distance, is when the height is zero.
That is, when it returns to the height of its starting point, which would be 4.5 metres.
4.5 = -0.04x^2+5.8x+4.9
0.04x^2 - 5.8x - 0.4 = 0
using the quadratic formula:
x = (5.8 ± √(33.64 - 4(.04)(-0.4)) / .08
= ...
I will let you do the button-pushing
You will get 2 answers, obvious you would reject the negative value of x
Let me know what you get.
I have a copy error, I don't know why I copied 4.9 as 4.5, so my equation should have
been
4.9 = -0.04x^2+5.8x+4.9
this makes it even easier:
.04x^2 - 5.8x = 0
x(.04x - 5.8) = 0
x = 0 or x = 5.8/.04 = .... , not one of your answers
I bet they did:
0 = 4.9 = -0.04x^2+5.8x+4.9
.04x^2 - 5.8x - 4.9 = 0
then x = 145.84 , which is one of their answers, but should not be according to the wording
Of course, I'll be happy to help you!
To find the horizontal distance the rocket will land, we need to determine the value of x when the rocket hits the ground. Since the rocket hits the ground, the height y must be 0.
We can set the equation equal to 0 and solve for x:
0 = -0.04x^2 + 5.8x + 4.9
Now, we have a quadratic equation. To solve it, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = -0.04, b = 5.8, and c = 4.9.
Substituting the values into the quadratic formula:
x = (-(5.8) ± √((5.8)^2 - 4(-0.04)(4.9))) / (2(-0.04))
Now, we can simply calculate this expression using a calculator to find the values of x. By plugging these values into the quadratic formula, we get:
x ≈ -0.21 or x ≈ 27.71
Since the horizontal distance cannot be negative, we can discard the negative solution. Therefore, the rocket will land approximately 27.71 meters horizontally from its starting point.
Good luck with your assignment! Let me know if there's anything else I can help with.