Chasity W.

asked • 11/14/12

find an equation that describes the path of a kick that covers a horizontal distance of 20 yards and reaches of maximum height of 9 yards

i really need help with my math. sos

3 Answers By Expert Tutors

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Alex O. answered • 11/14/12

Tutor
New to Wyzant

My name is Alex and I am a junior in petroleum engineering at UL.

Michael B.

It's not at all clear how you are solving for your 'c' variable...   You start with an inverted parabola y=-x2 + 9, and somehow you end up with "c*-10^2+9=0"??  I'm not sure what you are doing there.  What is "c*"?  Where did 10^2 come from?

Also, your final equation doesn't satisfy the requirement of a maximum height of 9, but I think it's because you just typed the wrong number...   I think your final equation (the one that is bolded) was meant to be:

y = [-.09(x-10)2]+9

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11/15/12

Robert J.

I like the original ideas. It is clear to me.

The ideas is to use transformation to get the answer.

You start with y = ax^2+9, which means the axis of symmetry is y-axis.

When x = 10, y = 0, thus you get a = -9/100.

So, y = -(9/100)x^2 + 9 <==This is a correct answer.

If you want to start with x = 0, then you need to shift parabola to the right by 10.

Therefore, the answer becomes y = -(9/100)(x-10)^2 + 9, which is equivalent to my answer y = -(9/100)(x)(x-20).

 

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11/15/12

Alex O.

Yes you are right. I meant to put + 9. Your bolded equation is what I meant.

The part with getting "c" was unclear so I will try again.

We have to strech the graph to get the desired distance of the kick. To strech a graph to have to multiply x by some factor. I just chose "c" to represent that factor. You could have picked any letter to represent the factor.

To find it I plugged in the desired values of y=0 and x=10 into the equation. That is where the 102 came from. I think that is where I lost you. I'll try to explain it better.

I chose these values because my max height, at that time, was at x=0. We know that the max height will be reached when half the total distance is travelled. So I need 10 units(half the distance) before and after the max height. Y=0 represents the kick hitting the ground. So after the max height, y(the height) should be zero at x=10(this would represent the second half of total travel). This would give ten units on each side of the max height and 20 units in between the two instances when y=0(start to finish). 

So now that I have royaly confused you again, we can find c by plugging in y=0 at x=10.

y=(c)*-x2+9  ...  again c is just that factor we are trying to find.   Now plug in 0 for y and 10 for x and we get... 0=c*-102+9 ... solve for c and get... c=9/100=0.09

Now we plug c back into the original equation.. y=0.09*-x2+9.. This equation has a max height of 9 and total distance of 20. But the max height is still at the x=0. From here we just shift the graph to the right to get the equation to start at the origin.

I hope this helps!

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11/15/12

Alex O.

Sorry Robert it is the same answer. I commented before you changed the typo.

I actually didn't remember how I was taught to do this so I just went to the basics of shifting and scaling.

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11/15/12

Michael B.

Alex, I think it's just your notation that is confusing...  the *- in your equations is very unconventional.  But I think I follow now..   I think that when you write:

c*-x2+9    (and similar in multiple places)

What you really mean is 

(c)(-1)(x2)+9

or better:   -c*x2+9

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11/15/12

Robert J. answered • 11/14/12

Tutor
4.6 (13)

Certified High School AP Calculus and Physics Teacher

Michael B.

This is a good method, but you confused something... Maximum height should be 9, not 20.  Thus:

9 = a(10)(-10)

a = -0.09

y = -0.09(x)(x-20)

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11/15/12

Robert J.

It's a typo. I changed it. Thank you for indicating the error.

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11/15/12

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