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heathernelly
 3 years ago
can someone please
heathernelly
 3 years ago
can someone please

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him1618
 3 years ago
Best ResponseYou've already chosen the best response.1have u done parametric equations?

him1618
 3 years ago
Best ResponseYou've already chosen the best response.1and do you know cartesian coordinates?

him1618
 3 years ago
Best ResponseYou've already chosen the best response.1lets take the feet of the archer to be (0,0)

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0these are the points i got (0,1.39) (18,8) (45,0) :/

him1618
 3 years ago
Best ResponseYou've already chosen the best response.1very good..now we need an equation which relates x and y dont we?

him1618
 3 years ago
Best ResponseYou've already chosen the best response.1lets say he launches the arrow with a horizontal velocity of v..the vertical velocity initaially is zero

him1618
 3 years ago
Best ResponseYou've already chosen the best response.1can u tell me the xcoordinate at any point in time?

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

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

him1618
 3 years ago
Best ResponseYou've already chosen the best response.1similarly, 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

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

phi
 3 years ago
Best ResponseYou've already chosen the best response.2I 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?

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

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

phi
 3 years ago
Best ResponseYou've already chosen the best response.2Here is the other post http://openstudy.com/users/phi#/updates/507e201be4b0919a3cf31396

phi
 3 years ago
Best ResponseYou've already chosen the best response.2The 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?

phi
 3 years ago
Best ResponseYou've already chosen the best response.2so you use the picture to find the height of A above the ground dw:1350594969329:dw

phi
 3 years ago
Best ResponseYou've already chosen the best response.2In 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.

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

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

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

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

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0where did you get the 4.9 from?

phi
 3 years ago
Best ResponseYou've already chosen the best response.2I used g= 9.8 m/s^2 in 1/2 g t^2

phi
 3 years ago
Best ResponseYou've already chosen the best response.2you have to redo the numbers. And I notice a typo... but it all changes anyway...

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0is the 1.035 the mistake?

phi
 3 years ago
Best ResponseYou've already chosen the best response.2yes, that should have been 1.35 but now it should be 1.26

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0yeahh :) 18tanθ + 0.04 where did the 0.04 come from?

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0nevermind, i know now (:

phi
 3 years ago
Best ResponseYou've already chosen the best response.2let me know what you get, and I'll double check it.

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0what does that zero thing mean after tan ? −18tanθ

phi
 3 years ago
Best ResponseYou've already chosen the best response.2which equation number are you looking at?

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.00 =−18tanθ − 0.04 /18^2 45^2 + 45tanθ +1.39

phi
 3 years ago
Best ResponseYou've already chosen the best response.2I 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
 3 years ago
Best ResponseYou've already chosen the best response.2I labeled this equation (6) so you can refer to it.

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0i mean like how did you get this  tanθ =19/1125 = 0.0168

phi
 3 years ago
Best ResponseYou've already chosen the best response.2solve for tanθ it is a bit messy, and I did not bother to write it all down. It is algebra. You do know algebra :)

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0haha 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
 3 years ago
Best ResponseYou've already chosen the best response.2are you working with the new numbers, or just working through the numbers in this old post?

phi
 3 years ago
Best ResponseYou've already chosen the best response.2what equation do you have now?

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0i have 0 = 18tan  0.13 / 18^2? :/

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

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0yeaah i have it, i forgot to say that

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0i actually have for the rest 45^2 + 45 tan + 1.39

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

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

phi
 3 years ago
Best ResponseYou've already chosen the best response.2get 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

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0yes, thank you soo soo much for the help :)

phi
 3 years ago
Best ResponseYou've already chosen the best response.2post your results. I'll check back laster.

TiffanyLee3
 3 years ago
Best ResponseYou've already chosen the best response.0Has anyone figured this out?

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0not yet, still trying to figure it out :)

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.2how did you get 0.01283... ?

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

phi
 3 years ago
Best ResponseYou've already chosen the best response.2you 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 \[\tan(\theta) = \frac{77}{6000}\cdot \frac{2}{3}\]

phi
 3 years ago
Best ResponseYou've already chosen the best response.2so 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
 3 years ago
Best ResponseYou've already chosen the best response.2I think you missed a decimal place

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

phi
 3 years ago
Best 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

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0i just got kinda confused :(

phi
 3 years ago
Best ResponseYou've already chosen the best response.2Yes, 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
 3 years ago
Best ResponseYou've already chosen the best response.2just get out your calculator or type it in google: (18*77/9000 0.13)/(18^2)=

phi
 3 years ago
Best ResponseYou've already chosen the best response.2oooh, Detroit swept the Yankees 4 games to 0. Detroit is off to the World Series

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0haha, i dont watch that ..but woo for them! :D

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

phi
 3 years ago
Best ResponseYou've already chosen the best response.2so now you have the full equation.

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0yay ..so do i use the 0.0085 & that?:)

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0omg, i would of never done this without you

phi
 3 years ago
Best ResponseYou've already chosen the best response.2you still have to find the max height of y

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0yeahh , so i did that next step ..which i got 3.725

phi
 3 years ago
Best ResponseYou've already chosen the best response.2The 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?

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0how did i get 3.725? :/

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

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0yes i used b/2a to get the answer 3.7

phi
 3 years ago
Best ResponseYou've already chosen the best response.2maybe you should type out the equation y = 0.0008765 x^2 + 0.0085 x + 1.39

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0yess, i wrote that down on my paper :) haha

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0but i had it the wrong way around :/ oh

phi
 3 years ago
Best ResponseYou've already chosen the best response.2yes, I notice you are not careful enough.

heathernelly
 3 years ago
Best ResponseYou've already chosen the best response.0im confused on what to do next

phi
 3 years ago
Best ResponseYou've already chosen the best response.2y = 0.0008765 x^2 + 0.0085 x + 1.39 type in google  0.0085/(2*0.0008765)=

phi
 3 years ago
Best ResponseYou've already chosen the best response.2much better. that is the x value where the parabola peaks. what is the y value at that x value?

phi
 3 years ago
Best ResponseYou've already chosen the best response.2type this in google 0.0008765*4.8488^2 + 0.0085*4.8488 + 1.39=
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