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BreezeFlow
Group Title
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:
 one year ago
 one year ago
BreezeFlow Group Title
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:
 one year ago
 one year ago

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theEric Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
this question has been posted before. there's not enough info here to reach a conclusion without making an assumption...
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
Can you see the picture?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
"From top to bottom the whole target stand is supposed to be 1.5 meters." where does it say that??
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
then it's easy to solve.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Haha, that might be a necessary piece of information! It makes point A known! :)
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
Oh good it can be solved :D
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
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! ?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
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...
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
IS 1.1 the height of A?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
OS crashed the pdf page... what were the measurements...
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
1.5  the distance from the top... whatever it was something like 4cm * 4 + 8cm?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
The height was 1.5m, and the point was 24 cm down.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
.24m down.
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
ah ok so (18, 1.26) is the second point...
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
Do I put the points into the quadratic equation form?
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
When I was working on this earlier I put c as being 1.39 for one of the equations.
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
I'd use... y= a(xh)^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...
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
1.26 = a(18)^2 +b18 +1.39 0 = a(45)^2 +b(45) +1.39
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
a is obviously going to be negative..
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
Do i solve it like linear systems?
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
Can I work this out and you guys tell me if i did it right?
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
Ok, thank you I'll get working on it.
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
What should I do instead?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
I solved .13 = 18^2a +18b for b and subsed in to the other equation a~ 8.765E4 b~ .008555
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
To find the maximum height do I use the formula b/2a?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
that's the x location of the max height... you plug that into the equation to find the actual height...
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
What's the equation for height. Oh and with the number for a should i type e4 in my calculator?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
we just found the equation for height...
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
We use the second equation?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
the equation of the parabola... we were using the points... to find the equation of the parabola to find its vertex....
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
the vertex is the location of the max height
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
dw:1350531626084:dw
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
Does that mean the ycoordinate of the vertex is the max height?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
(x,y) = (x location of max height, max height)
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
Do i already know the vertex? Was it one of the points from earlier?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
yeah it was (45,0)
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
you already knew the answer before you did the problem, so this was all a huge waste of time.
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
I still don't get how to arrive to the answer and I need to show the work
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
find a and b y= ax^2 +bx +c find the vertex
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
If the vertex is supposed to be (45,0) wouldn't that make the maximum height 0? o_o
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
(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
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
When I try to find the height I get 4.88025002715, but I know that can't be write.
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
it's not.
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
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 ycoordinate.
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
plug x =b/2a into the equation of the parabola
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
The ax^2+bx+c type equation?
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
alternately, calculate (4ac b^2) / 4a your choice.
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
Ok I'll try that
 one year ago

BreezeFlow Group TitleBest ResponseYou've already chosen the best response.0
I got 1.3691, but I probably miscalculated somewhere unless I'm supposed to round it to 1.4
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
it's 1.41
 one year ago

Algebraic! Group TitleBest ResponseYou've already chosen the best response.1
1.410876 using the rough values for a and b that I gave
 one year ago
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