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Robin's question at Yahoo! Answers regarding the height of a rocket over sloping ground

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MarkFL

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Feb 24, 2012
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Here is the question:

PRECAL INVERSE FUNCTION PROBLEM PLEASE HELP?


Clovis is standing at the edge of a cliff, which slopes 4 feet downward from him for every 1 horizontal foot. He launches a small model rocket from where he is standing.

With the origin of the coordinate system located where he is standing, and the x-axis extending horizontally, the path of the rocket is described by the formula y= -2x^2 + 120x.

a.) Give a function h = f(x) relating the height h of the rocket above the sloping ground to its x-coordinate.

0 = -2x^2 +124

b.) Find the maximum height of the rocket above the sloping ground. What is its x-coordinate when it is at its maximum height?

MAX: 31, 1922

c.) Clovis measures its height h of the rocket above the sloping ground while it is going up. Give a function x = g(h) relating the x-coordinate of the rocket to h.

The maximum I get for the rocket is (31, 1922)
I am pretty sure it is right because the answer in the back of the book is 31 - 1/2 sqrt3844 - 2h
so the 31 is probably correct!
I just need the answer to C (obviously), I'd really appreciate it if you could guide me through it rather than just answers.
Please no use of Calculus terms, I am in Precalc.
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

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Feb 24, 2012
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Hello Robin,

We are given the trajectory of the rocket as:

\(\displaystyle h(x)=-2x^2+120x\)

And we may model the surface of the ground by observing it passes through the origin of our coordinate system and has a slope of $-4$. Thus the "ground" function is:

\(\displaystyle g(x)=-4x\)

a.) Give a function h = f(x) relating the height h of the rocket above the sloping ground to its x-coordinate.

The height of the rocket above the ground at $x$ is the difference between the two functions:

\(\displaystyle f(x)=h(x)-g(x)=\left(-2x^2+120x \right)-(-4x)=-2x^2+124x=2x(62-x)\)

I think you meant to give this but simply made a typo and left off the variable of the linear term.

b.) Find the maximum height of the rocket above the sloping ground. What is its x-coordinate when it is at its maximum height?

From part a) we see the two roots of $f(x)$ are:

\(\displaystyle x=0,\,62\)

Now, we know the axis of symmetry, along which the vertex lies, will be midway between the roots. Using the mid-point formula, we then find the axis of symmetry is:

\(\displaystyle x=\frac{0+62}{2}=31\)

The height at the value of $x$ is:

\(\displaystyle f(31)=2\cdot31(62-31)=2\cdot31^2=1922\)

Hence:

\(\displaystyle f_{\max}=f(31)=1922\)

We have found that the rocket reaches a maximum height of 1922 ft at a horizontal distance of 31 ft from the origin.

This agrees with what you found. (Clapping)

c.) Clovis measures its height h of the rocket above the sloping ground while it is going up. Give a function x = g(h) relating the x-coordinate of the rocket to h.

Recall we found:

\(\displaystyle f(x)=-2x^2+124x\)

Arrange in standard quadratic form:

\(\displaystyle 2x^2-124x+f=0\)

Application of the quadratic formula yields:

\(\displaystyle x=\frac{124\pm\sqrt{(-124)^2-4(2)(f)}}{2(2)}=\frac{62\pm\sqrt{3844-2f}}{2}\)

Now, since $f$ is not a one-to-one function, we must observe that the relevant domain of $f$ is \(\displaystyle 0\le x\le62\) and so we must take the branch:

\(\displaystyle x=\frac{62-\sqrt{3844-2f}}{2}=31-\frac{\sqrt{3844-2f}}{2}\)