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heathernelly
can someone please
have u done parametric equations?
and do you know cartesian coordinates?
lets take the feet of the archer to be (0,0)
these are the points i got (0,1.39) (18,8) (45,0) :/
very good..now we need an equation which relates x and y dont we?
lets say he launches the arrow with a horizontal velocity of v..the vertical velocity initaially is zero
can u tell me the x-coordinate at any point in time?
hmm, i don't know :/ ? im not very good at math.. ha
distance is speed x time..since there is no acceleration in the x-direction, the speed always remains v..so x = vt
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
my teacher said that i would have to use quadratic equation to do this problem thou ..does this lead to that orr? :/
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?
yeah, there's a diagram , i forgot what their called .. but next to the bulls eye it says A (8cm)
my teacher also said The distance from the top edge of the archery target to the ground is 1.5 meters
Here is the other post http://openstudy.com/users/phi#/updates/507e201be4b0919a3cf31396
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?
so you use the picture to find the height of A above the ground |dw:1350594969329:dw|
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.
I did not explain where the equations come from (it's physics), so it is just showing the math
so all i need to do now would be to use the equation ax^2+bx+c ? & then put the #'s in ?
did you find my write-up (it's an attachment in the 2nd to last post)
yes got it, you are awesomee! <33 i don't know what i would of donee. haha
where did you get the -4.9 from?
I used g= 9.8 m/s^2 in 1/2 g t^2
you have to re-do the numbers. And I notice a typo... but it all changes anyway...
is the 1.035 the mistake?
yes, that should have been 1.35 but now it should be 1.26
yeahh :) 18tanθ + 0.04 where did the 0.04 come from?
nevermind, i know now (:
let me know what you get, and I'll double check it.
what does that zero thing mean after tan ? −18tanθ
which equation number are you looking at?
0 =−18tanθ − 0.04 /18^2 45^2 + 45tanθ +1.39
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
I labeled this equation (6) so you can refer to it.
i mean like how did you get this - tanθ =19/1125 = 0.0168
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 :)
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 :/
are you working with the new numbers, or just working through the numbers in this old post?
what equation do you have now?
i have 0 = 18tan - 0.13 / 18^2? :/
what about the rest? I would expect to see \[ y= \frac{18 \tan(\theta)-0.13}{18^2}x^2 + x \tan(\theta)+1.39\]
yeaah i have it, i forgot to say that
i actually have for the rest 45^2 + 45 tan + 1.39
yes, after you sub in (45,0) you get \[ 0= \frac{18\tan(\theta)-0.13}{18^2}45^2+45 \tan(\theta)+1.39\]
yeah :) but then i get confused after that bit because i dont know how you got 19/1125
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
yes, thank you soo soo much for the help :)
post your results. I'll check back laster.
Has anyone figured this out?
not yet, still trying to figure it out :)
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 :/
how did you get 0.01283... ?
from 0=(1-5/2)tan - 0.13/18 x 5/2 + 1.39/45 :/
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 \[\tan(\theta) = \frac{77}{6000}\cdot \frac{2}{3}\]
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
I think you missed a decimal place
yeah, haha ..sorry 0.0085 ..so i would use that & the number i said up there^^?
\[ y= \frac{-18 \tan(\theta)-0.13}{18^2}x^2 + x \tan(\theta)+1.39 \] you need to find the coeff for the x^2 term. but you know tan so you can simplify
i just got kinda confused :(
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
just get out your calculator or type it in google: (-18*77/9000 -0.13)/(18^2)=
oooh, Detroit swept the Yankees 4 games to 0. Detroit is off to the World Series
haha, i dont watch that ..but woo for them! :D
you have to be more careful. put in a decimal point, and it is not 9 for the first digit
so now you have the full equation.
yay ..so do i use the 0.0085 & that?:)
omg, i would of never done this without you
you still have to find the max height of y
yeahh , so i did that next step ..which i got -3.725
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?
how did i get -3.725? :/
if you have y = ax^2 +bx+c you should use x_max= -b/(2a)
yes i used -b/2a to get the answer -3.7
maybe you should type out the equation y = -0.0008765 x^2 + 0.0085 x + 1.39
yess, i wrote that down on my paper :) haha
but i had it the wrong way around :/ oh
yes, I notice you are not careful enough.
im confused on what to do next
y = -0.0008765 x^2 + 0.0085 x + 1.39 type in google - 0.0085/(2*-0.0008765)=
much better. that is the x value where the parabola peaks. what is the y value at that x value?
type this in google -0.0008765*4.8488^2 + 0.0085*4.8488 + 1.39=