heathernelly 3 years ago can someone please

1. him1618

have u done parametric equations?

2. him1618

and do you know cartesian coordinates?

3. heathernelly

no:/

4. him1618

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

5. heathernelly

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

6. him1618

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

7. heathernelly

yeaah

8. him1618

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

9. him1618

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

10. heathernelly

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

11. him1618

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

12. him1618

understand?

13. heathernelly

yeah,, kindaa

14. him1618

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

15. heathernelly

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

16. 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?

17. heathernelly

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

18. heathernelly

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

19. phi

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

20. 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?

21. heathernelly

yess, it is

22. phi

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

23. 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.

24. phi

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

25. heathernelly

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

26. phi

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

27. heathernelly

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

28. heathernelly

where did you get the -4.9 from?

29. phi

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

30. phi

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

31. heathernelly

is the 1.035 the mistake?

32. phi

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

33. heathernelly

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

34. heathernelly

nevermind, i know now (:

35. phi

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

36. heathernelly

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

37. phi

which equation number are you looking at?

38. heathernelly

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

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

40. phi

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

41. heathernelly

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

42. 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 :)

43. heathernelly

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 :/

44. phi

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

45. heathernelly

new numbers, haha

46. phi

what equation do you have now?

47. heathernelly

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

48. phi

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

49. heathernelly

yeaah i have it, i forgot to say that

50. heathernelly

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

51. phi

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

52. heathernelly

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

53. 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

54. heathernelly

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

55. phi

post your results. I'll check back laster.

56. heathernelly

will do :)

57. TiffanyLee3

Has anyone figured this out?

58. heathernelly

not yet, still trying to figure it out :)

59. heathernelly

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 :/

60. phi

how did you get 0.01283... ?

61. heathernelly

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

62. 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 $\tan(\theta) = \frac{77}{6000}\cdot \frac{2}{3}$

63. heathernelly

so 0.085

64. 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

65. phi

I think you missed a decimal place

66. heathernelly

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

67. 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

68. heathernelly

i just got kinda confused :(

69. 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

70. phi

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

71. heathernelly

-0009765

72. phi

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

73. heathernelly

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

74. phi

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

75. heathernelly

8 i meaan (:

76. heathernelly

i mean -0.0008765

77. phi

so now you have the full equation.

78. heathernelly

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

79. phi

yes

80. heathernelly

omg, i would of never done this without you

81. phi

you still have to find the max height of y

82. heathernelly

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

83. 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?

84. phi

How did you find x?

85. heathernelly

how did i get -3.725? :/

86. phi

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

87. heathernelly

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

88. heathernelly

25

89. phi

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

90. heathernelly

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

91. heathernelly

but i had it the wrong way around :/ oh

92. phi

yes, I notice you are not careful enough.

93. heathernelly

im confused on what to do next

94. phi

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

95. heathernelly

4.8488

96. phi

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

97. phi

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

98. heathernelly

1.41