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him1618 Group TitleBest ResponseYou've already chosen the best response.1
have u done parametric equations?
 one year ago

him1618 Group TitleBest ResponseYou've already chosen the best response.1
and do you know cartesian coordinates?
 one year ago

him1618 Group TitleBest ResponseYou've already chosen the best response.1
lets take the feet of the archer to be (0,0)
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
these are the points i got (0,1.39) (18,8) (45,0) :/
 one year ago

him1618 Group TitleBest ResponseYou've already chosen the best response.1
very good..now we need an equation which relates x and y dont we?
 one year ago

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

him1618 Group TitleBest ResponseYou've already chosen the best response.1
lets say he launches the arrow with a horizontal velocity of v..the vertical velocity initaially is zero
 one year ago

him1618 Group TitleBest ResponseYou've already chosen the best response.1
can u tell me the xcoordinate at any point in time?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
hmm, i don't know :/ ? im not very good at math.. ha
 one year ago

him1618 Group TitleBest ResponseYou've already chosen the best response.1
distance is speed x time..since there is no acceleration in the xdirection, the speed always remains v..so x = vt
 one year ago

him1618 Group TitleBest ResponseYou've already chosen the best response.1
understand?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yeah,, kindaa
 one year ago

him1618 Group TitleBest ResponseYou've already chosen the best response.1
similarly, there is a downward acceleration of 10 m/s^2 in the ydirection.. 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
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
my teacher said that i would have to use quadratic equation to do this problem thou ..does this lead to that orr? :/
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yeah, there's a diagram , i forgot what their called .. but next to the bulls eye it says A (8cm)
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
my teacher also said The distance from the top edge of the archery target to the ground is 1.5 meters
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
Here is the other post http://openstudy.com/users/phi#/updates/507e201be4b0919a3cf31396
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yess, it is
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
so you use the picture to find the height of A above the ground dw:1350594969329:dw
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
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 redone.
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
I did not explain where the equations come from (it's physics), so it is just showing the math
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
so all i need to do now would be to use the equation ax^2+bx+c ? & then put the #'s in ?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
did you find my writeup (it's an attachment in the 2nd to last post)
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yes got it, you are awesomee! <33 i don't know what i would of donee. haha
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
where did you get the 4.9 from?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
I used g= 9.8 m/s^2 in 1/2 g t^2
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
you have to redo the numbers. And I notice a typo... but it all changes anyway...
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
is the 1.035 the mistake?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
yes, that should have been 1.35 but now it should be 1.26
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yeahh :) 18tanθ + 0.04 where did the 0.04 come from?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
nevermind, i know now (:
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
let me know what you get, and I'll double check it.
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
what does that zero thing mean after tan ? −18tanθ
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
which equation number are you looking at?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
0 =−18tanθ − 0.04 /18^2 45^2 + 45tanθ +1.39
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
I labeled this equation (6) so you can refer to it.
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
i mean like how did you get this  tanθ =19/1125 = 0.0168
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
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 :)
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
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 :/
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
are you working with the new numbers, or just working through the numbers in this old post?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
new numbers, haha
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
what equation do you have now?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
i have 0 = 18tan  0.13 / 18^2? :/
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
what about the rest? I would expect to see \[ y= \frac{18 \tan(\theta)0.13}{18^2}x^2 + x \tan(\theta)+1.39\]
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yeaah i have it, i forgot to say that
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
i actually have for the rest 45^2 + 45 tan + 1.39
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
yes, after you sub in (45,0) you get \[ 0= \frac{18\tan(\theta)0.13}{18^2}45^2+45 \tan(\theta)+1.39\]
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yeah :) but then i get confused after that bit because i dont know how you got 19/1125
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
get out a piece of paper and use the distributive property and rewrite 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
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yes, thank you soo soo much for the help :)
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
post your results. I'll check back laster.
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
will do :)
 one year ago

TiffanyLee3 Group TitleBest ResponseYou've already chosen the best response.0
Has anyone figured this out?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
not yet, still trying to figure it out :)
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
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 :/
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
how did you get 0.01283... ?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
from 0=(15/2)tan  0.13/18 x 5/2 + 1.39/45 :/
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
you start with \[ 0= (1\frac{5}{2})\tan(\theta) \frac{0.13}{18}\cdot \frac{5}{2}+\frac{1.39}{45}\] 15/2 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 \[\tan(\theta) = \frac{77}{6000}\cdot \frac{2}{3}\]
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
so 0.085
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
I think you missed a decimal place
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yeah, haha ..sorry 0.0085 ..so i would use that & the number i said up there^^?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
\[ y= \frac{18 \tan(\theta)0.13}{18^2}x^2 + x \tan(\theta)+1.39 \] you need to find the coeff for the x^2 term. but you know tan so you can simplify
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
i just got kinda confused :(
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
just get out your calculator or type it in google: (18*77/9000 0.13)/(18^2)=
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
0009765
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
oooh, Detroit swept the Yankees 4 games to 0. Detroit is off to the World Series
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
haha, i dont watch that ..but woo for them! :D
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
you have to be more careful. put in a decimal point, and it is not 9 for the first digit
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
8 i meaan (:
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
i mean 0.0008765
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
so now you have the full equation.
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yay ..so do i use the 0.0085 & that?:)
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
omg, i would of never done this without you
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
you still have to find the max height of y
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yeahh , so i did that next step ..which i got 3.725
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
How did you find x?
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
how did i get 3.725? :/
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
if you have y = ax^2 +bx+c you should use x_max= b/(2a)
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yes i used b/2a to get the answer 3.7
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
maybe you should type out the equation y = 0.0008765 x^2 + 0.0085 x + 1.39
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
yess, i wrote that down on my paper :) haha
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
but i had it the wrong way around :/ oh
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
yes, I notice you are not careful enough.
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
im confused on what to do next
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
y = 0.0008765 x^2 + 0.0085 x + 1.39 type in google  0.0085/(2*0.0008765)=
 one year ago

heathernelly Group TitleBest ResponseYou've already chosen the best response.0
4.8488
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
much better. that is the x value where the parabola peaks. what is the y value at that x value?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.2
type this in google 0.0008765*4.8488^2 + 0.0085*4.8488 + 1.39=
 one year ago
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