anonymous
  • anonymous
I would really appreciate if someone could help me with this problem. It relates parabolas and archery. I've been trying to figure it out for about two hours now and I can't seem to get to the right answer, which I already know. Here's a link so you can see the picture with the problem: http://mrswhaples.cmswiki.wikispaces.net/file/view/Algebra+II+chapter+4+solutions.pdf The problem I need help with is on the second last page under Activity 2. Every time I attempt to get to the right answer, I am way off. D:
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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theEric
  • theEric
Hi! I want to help, but I'm not quite sure how this is working. So maybe we can work together to get to the right answer. You can tell me what you did and I'll try to understand it, or we can start from the beginning. Is it assumed that the archer shoots horizontally? And how high is the target? Are we able to know that? Sometimes I think we'd need to know things that we really don't. Variables cancel out, sometimes.. Also, I'm sorry you've spent so much time on this so far!
anonymous
  • anonymous
Thank you for being willing to help. So far I've been using three points to try to find an answer: (0, 1.39) (18,40) and (45,0). From top to bottom the whole target stand is supposed to be 1.5 meters. And the arrow is being shot at 1.39 meters.
anonymous
  • anonymous
this question has been posted before. there's not enough info here to reach a conclusion without making an assumption...

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anonymous
  • anonymous
Can you see the picture?
anonymous
  • anonymous
"From top to bottom the whole target stand is supposed to be 1.5 meters." where does it say that??
anonymous
  • anonymous
On my homework sheet it says that. It's supposed to be a tip. Use the following piece of information to help you solve Activity 2: The distance from the top edge of the archery target to the ground is 1.5 meters
anonymous
  • anonymous
then it's easy to solve.
theEric
  • theEric
Haha, that might be a necessary piece of information! It makes point A known! :)
anonymous
  • anonymous
Oh good it can be solved :D
theEric
  • theEric
And I was silly to think the shot was horizontal - the answer would be shoulder height :P Do you want to take it away, @Algebraic! ?
anonymous
  • anonymous
not really... you have three points now. (0, 1.39) (18, 1.1) and (45,0) should not be a problem to find the vertex of that parabola...
anonymous
  • anonymous
IS 1.1 the height of A?
anonymous
  • anonymous
OS crashed the pdf page... what were the measurements...
anonymous
  • anonymous
1.5 - the distance from the top... whatever it was something like 4cm * 4 + 8cm?
theEric
  • theEric
The height was 1.5m, and the point was 24 cm down.
theEric
  • theEric
.24m down.
anonymous
  • anonymous
Here's the typed out problem An archer releases an arrow from a shoulder height of 1.39 m. When the arrow hits the target 18 m away, it hits point A. When the target is removed, the arrow lands 45 m away. Find the maximum height of the arrow along its parabolic path. 4 cm 4 cm 4 cm 4 cm 8 cm
anonymous
  • anonymous
ah ok so (18, 1.26) is the second point...
anonymous
  • anonymous
Do I put the points into the quadratic equation form?
anonymous
  • anonymous
When I was working on this earlier I put c as being 1.39 for one of the equations.
anonymous
  • anonymous
I'd use... y= a(x-h)^2 +k for ease... but yeah you can use y = ax^2 +bx +c with c=1.39m if you like... doesn't really matter...
anonymous
  • anonymous
1.26 = a(18)^2 +b18 +1.39 0 = a(45)^2 +b(45) +1.39
anonymous
  • anonymous
a is obviously going to be negative..
anonymous
  • anonymous
Do i solve it like linear systems?
anonymous
  • anonymous
yep
anonymous
  • anonymous
Can I work this out and you guys tell me if i did it right?
anonymous
  • anonymous
sure
anonymous
  • anonymous
Ok, thank you I'll get working on it.
anonymous
  • anonymous
Ok so far I simplified the expressions to 324a + 18b +1.39= 1.26 and 2025a+45b+1.39=0 Do I then subtract 1.39 from both sides and use the method of multiplying on equation by a number so that the coefficients of one variable will be opposites?
anonymous
  • anonymous
From subtracting 1.39 from both sides I got 324a + 18b = -0.13 and 2025a+45b=-1.39 Now, should I multiply one equation like 324a + 18b = -0.13 by -2.5 so the coefficient of b will be -45?
anonymous
  • anonymous
no.
anonymous
  • anonymous
What should I do instead?
anonymous
  • anonymous
I solved -.13 = 18^2a +18b for b and subsed in to the other equation a~ -8.765E-4 b~ .008555
anonymous
  • anonymous
To find the maximum height do I use the formula -b/2a?
anonymous
  • anonymous
that's the x location of the max height... you plug that into the equation to find the actual height...
anonymous
  • anonymous
What's the equation for height. Oh and with the number for a should i type e-4 in my calculator?
anonymous
  • anonymous
10^-4
anonymous
  • anonymous
we just found the equation for height...
anonymous
  • anonymous
We use the second equation?
anonymous
  • anonymous
the equation of the parabola... we were using the points... to find the equation of the parabola to find its vertex....
anonymous
  • anonymous
I have to find the vertex before I find the height? Sorry, if I sound stupid, I've been really confused with this problem for awhile. XP
anonymous
  • anonymous
the vertex is the location of the max height
anonymous
  • anonymous
|dw:1350531626084:dw|
anonymous
  • anonymous
Does that mean the y-coordinate of the vertex is the max height?
anonymous
  • anonymous
(x,y) = (x location of max height, max height)
anonymous
  • anonymous
yes
anonymous
  • anonymous
Do i already know the vertex? Was it one of the points from earlier?
anonymous
  • anonymous
yeah it was (45,0)
anonymous
  • anonymous
you already knew the answer before you did the problem, so this was all a huge waste of time.
anonymous
  • anonymous
I still don't get how to arrive to the answer and I need to show the work
anonymous
  • anonymous
find a and b y= ax^2 +bx +c find the vertex
anonymous
  • anonymous
If the vertex is supposed to be (45,0) wouldn't that make the maximum height 0? o_o
anonymous
  • anonymous
(0, 1.39) was a point so y= a(0)^2 +b(0) +c 1.39 =c (18, 1.26) was a point so 1.26 = a(18^2) + b(18) +1.39 (45,0) was a point so 0 = a(45)^2 +b(45)+ 1.39
anonymous
  • anonymous
When I try to find the height I get -4.88025002715, but I know that can't be write.
anonymous
  • anonymous
it's not.
anonymous
  • anonymous
I don't know how the answer is supposed to be 1.41 and I don't know how to find the vertex that has 1.41 as the y-coordinate.
anonymous
  • anonymous
plug x =-b/2a into the equation of the parabola
anonymous
  • anonymous
The ax^2+bx+c type equation?
anonymous
  • anonymous
alternately, calculate (4ac -b^2) / 4a your choice.
anonymous
  • anonymous
Ok I'll try that
anonymous
  • anonymous
I got 1.3691, but I probably miscalculated somewhere unless I'm supposed to round it to 1.4
anonymous
  • anonymous
it's 1.41
anonymous
  • anonymous
1.410876 using the rough values for a and b that I gave

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