can someone please

- anonymous

can someone please

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- anonymous

have u done parametric equations?

- anonymous

and do you know cartesian coordinates?

- anonymous

no:/

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## More answers

- anonymous

lets take the feet of the archer to be (0,0)

- anonymous

these are the points i got (0,1.39) (18,8) (45,0) :/

- anonymous

very good..now we need an equation which relates x and y dont we?

- anonymous

yeaah

- anonymous

lets say he launches the arrow with a horizontal velocity of v..the vertical velocity initaially is zero

- anonymous

can u tell me the x-coordinate at any point in time?

- anonymous

hmm, i don't know :/ ? im not very good at math.. ha

- anonymous

distance is speed x time..since there is no acceleration in the x-direction, the speed always remains v..so x = vt

- anonymous

understand?

- anonymous

yeah,, kindaa

- anonymous

similarly, there is a downward acceleration of -10 m/s^2 in the y-direction..
in the presence of accn, we have the distance covered as -0.5at^2, where a=accn
since we are starting at y=1.39 m, we remodel y as 1.39 - 0.5at^2, a= 10
therefore y= 1.39 - 5t^2

- anonymous

my teacher said that i would have to use quadratic equation to do this problem thou ..does this lead to that orr? :/

- phi

I posted an answer to this same question (but a different user posted it)
But the problem is I don't believe the height of A. You have it as 8 cm.
In the real world, this problem does not make sense. Are you sure about the height?

- anonymous

yeah, there's a diagram , i forgot what their called .. but next to the bulls eye it says A (8cm)

- anonymous

my teacher also said The distance from the top edge of the archery target to the ground is 1.5 meters

- phi

Here is the other post
http://openstudy.com/users/phi#/updates/507e201be4b0919a3cf31396

- phi

The distance from the top edge of the archery target to the ground is 1.5 meters
That is very important. Is the picture the same as in the post I just listed?

- anonymous

yess, it is

- phi

so you use the picture to find the height of A above the ground
|dw:1350594969329:dw|

- phi

In my post I just assumed A was at (18,1.35)
however, based on your new info, A is at (18,1.26)
so the numbers have to be re-done.

- phi

I did not explain where the equations come from (it's physics), so it is just showing the math

- anonymous

so all i need to do now would be to use the equation ax^2+bx+c ? & then put the #'s in ?

- phi

did you find my write-up (it's an attachment in the 2nd to last post)

- anonymous

yes got it, you are awesomee! <33 i don't know what i would of donee. haha

- anonymous

where did you get the -4.9 from?

- phi

I used g= 9.8 m/s^2 in
1/2 g t^2

- phi

you have to re-do the numbers. And I notice a typo... but it all changes anyway...

- anonymous

is the 1.035 the mistake?

- phi

yes, that should have been 1.35 but now it should be 1.26

- anonymous

yeahh :)
18tanθ + 0.04
where did the 0.04 come from?

- anonymous

nevermind, i know now (:

- phi

let me know what you get, and I'll double check it.

- anonymous

what does that zero thing mean after tan ? −18tanθ

- phi

which equation number are you looking at?

- anonymous

0 =−18tanθ − 0.04
/18^2
45^2 + 45tanθ +1.39

- phi

I used the point (45,0) in the equation (the arrow hits the ground y=0 45 meters away from the archer x=45
so that 0 is y

- phi

I labeled this equation (6) so you can refer to it.

- anonymous

i mean like how did you get this -
tanθ =19/1125
= 0.0168

- phi

solve for tanθ
it is a bit messy, and I did not bother to write it all down. It is algebra. You do know algebra :)

- anonymous

haha yeah, wasnt the best at it but i know what tan is ..so i would type in my calculator -18tan ___? what goes in the blank thou :/

- phi

are you working with the new numbers, or just working through the numbers in this old post?

- anonymous

new numbers, haha

- phi

what equation do you have now?

- anonymous

i have 0 = 18tan - 0.13 / 18^2? :/

- phi

what about the rest?
I would expect to see
\[ y= \frac{18 \tan(\theta)-0.13}{18^2}x^2 + x \tan(\theta)+1.39\]

- anonymous

yeaah i have it, i forgot to say that

- anonymous

i actually have for the rest 45^2 + 45 tan + 1.39

- phi

yes, after you sub in (45,0) you get
\[ 0= \frac{18\tan(\theta)-0.13}{18^2}45^2+45 \tan(\theta)+1.39\]

- anonymous

yeah :) but then i get confused after that bit because i dont know how you got 19/1125

- phi

get out a piece of paper and use the distributive property and re-write the equation
\[ 0= \frac{-45^2\tan(\theta)}{18} -\frac{0.13}{18^2}45^2+45 \tan(\theta)+1.39\]
I divided everything by 45:
\[ 0= -\frac{45}{18}\tan(\theta) -\frac{0.13}{18^2}45+ \tan(\theta)+\frac{1.39}{45}\]
the 45/18 simplifies to 5/2. I moved the tan terms together
\[ 0= (1-\frac{5}{2})\tan(\theta) -\frac{0.13}{18}\cdot \frac{5}{2}+\frac{1.39}{45}\]
you can finish I hope

- anonymous

yes, thank you soo soo much for the help :)

- phi

post your results. I'll check back laster.

- anonymous

will do :)

- anonymous

Has anyone figured this out?

- anonymous

not yet, still trying to figure it out :)

- anonymous

i figure out that it would be 0.0128333 ..so y = ____x^2 + 0.0128333+1.39
but i can't figure out what the other # is :/

- phi

how did you get 0.01283...
?

- anonymous

from 0=(1-5/2)tan - 0.13/18 x 5/2 + 1.39/45 :/

- phi

you start with
\[ 0= (1-\frac{5}{2})\tan(\theta) -\frac{0.13}{18}\cdot \frac{5}{2}+\frac{1.39}{45}\]
1-5/2 is 2/2 -5/2= -3/2
\[0=-\frac{3}{2}\tan(\theta) -\frac{0.13}{18}\cdot \frac{5}{2}+\frac{1.39}{45}\]
or
\[\frac{3}{2}\tan(\theta) = -\frac{0.13}{18}\cdot \frac{5}{2}+\frac{1.39}{45}\]
the stuff on the right simplifies to 77/6000 (which is 0.01283...)
\[\frac{3}{2}\tan(\theta) = \frac{77}{6000} \]
and
\[\tan(\theta) = \frac{77}{6000}\cdot \frac{2}{3}\]

- anonymous

so 0.085

- phi

so tanθ = 77/9000 = 0.008555...
that is your coefficient on x
you can find the coefficient of the x^2 term by replacing tanθ

- phi

I think you missed a decimal place

- anonymous

yeah, haha ..sorry 0.0085 ..so i would use that & the number i said up there^^?

- phi

\[ y= \frac{-18 \tan(\theta)-0.13}{18^2}x^2 + x \tan(\theta)+1.39 \]
you need to find the coeff for the x^2 term. but you know tan so you can simplify

- anonymous

i just got kinda confused :(

- phi

Yes, this is a complicated problem, and you have to be methodical.
If you look at my pdf, we got to equation (5). Then we sub in (45,0) and solved for tan= 77/9000
now you can replace tanθ in equation (5):
\[ y= \frac{-18 \tan(\theta)-0.13}{18^2}x^2 + x \tan(\theta)+1.39 \]
you will get a parabola with numbers for coefficients

- phi

just get out your calculator or type it in google:
(-18*77/9000 -0.13)/(18^2)=

- anonymous

-0009765

- phi

oooh, Detroit swept the Yankees 4 games to 0. Detroit is off to the World Series

- anonymous

haha, i dont watch that ..but woo for them! :D

- phi

you have to be more careful. put in a decimal point, and it is not 9 for the first digit

- anonymous

8 i meaan (:

- anonymous

i mean -0.0008765

- phi

so now you have the full equation.

- anonymous

yay ..so do i use the 0.0085 & that?:)

- phi

yes

- anonymous

omg, i would of never done this without you

- phi

you still have to find the max height of y

- anonymous

yeahh , so i did that next step ..which i got -3.725

- phi

The next step is to find x where the parabola peaks.
The way the problem is set up, the archer is at x=0 and the target at x=18
x= -3.7.. is behind the archer. So maybe you got the wrong sign?

- phi

How did you find x?

- anonymous

how did i get -3.725? :/

- phi

if you have
y = ax^2 +bx+c
you should use x_max= -b/(2a)

- anonymous

yes i used -b/2a to get the answer -3.7

- anonymous

25

- phi

maybe you should type out the equation
y = -0.0008765 x^2 + 0.0085 x + 1.39

- anonymous

yess, i wrote that down on my paper :) haha

- anonymous

but i had it the wrong way around :/ oh

- phi

yes, I notice you are not careful enough.

- anonymous

im confused on what to do next

- phi

y = -0.0008765 x^2 + 0.0085 x + 1.39
type in google
- 0.0085/(2*-0.0008765)=

- anonymous

4.8488

- phi

much better. that is the x value where the parabola peaks. what is the y value at that x value?

- phi

type this in google
-0.0008765*4.8488^2 + 0.0085*4.8488 + 1.39=

- anonymous

1.41

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