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BreezeFlow
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:
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!
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.
this question has been posted before. there's not enough info here to reach a conclusion without making an assumption...
Can you see the picture?
"From top to bottom the whole target stand is supposed to be 1.5 meters." where does it say that??
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
then it's easy to solve.
Haha, that might be a necessary piece of information! It makes point A known! :)
Oh good it can be solved :D
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! ?
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...
IS 1.1 the height of A?
OS crashed the pdf page... what were the measurements...
1.5 - the distance from the top... whatever it was something like 4cm * 4 + 8cm?
The height was 1.5m, and the point was 24 cm down.
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
ah ok so (18, 1.26) is the second point...
Do I put the points into the quadratic equation form?
When I was working on this earlier I put c as being 1.39 for one of the equations.
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...
1.26 = a(18)^2 +b18 +1.39 0 = a(45)^2 +b(45) +1.39
a is obviously going to be negative..
Do i solve it like linear systems?
Can I work this out and you guys tell me if i did it right?
Ok, thank you I'll get working on it.
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?
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?
What should I do instead?
I solved -.13 = 18^2a +18b for b and subsed in to the other equation a~ -8.765E-4 b~ .008555
To find the maximum height do I use the formula -b/2a?
that's the x location of the max height... you plug that into the equation to find the actual height...
What's the equation for height. Oh and with the number for a should i type e-4 in my calculator?
we just found the equation for height...
We use the second equation?
the equation of the parabola... we were using the points... to find the equation of the parabola to find its vertex....
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
the vertex is the location of the max height
|dw:1350531626084:dw|
Does that mean the y-coordinate of the vertex is the max height?
(x,y) = (x location of max height, max height)
Do i already know the vertex? Was it one of the points from earlier?
you already knew the answer before you did the problem, so this was all a huge waste of time.
I still don't get how to arrive to the answer and I need to show the work
find a and b y= ax^2 +bx +c find the vertex
If the vertex is supposed to be (45,0) wouldn't that make the maximum height 0? o_o
(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
When I try to find the height I get -4.88025002715, but I know that can't be write.
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.
plug x =-b/2a into the equation of the parabola
The ax^2+bx+c type equation?
alternately, calculate (4ac -b^2) / 4a your choice.
I got 1.3691, but I probably miscalculated somewhere unless I'm supposed to round it to 1.4
1.410876 using the rough values for a and b that I gave