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YumYum247

  • one year ago

elp please!!!!

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  1. YumYum247
    • one year ago
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    Can someone please check question number 41 for me plz?!?!?!!?!?

  2. YumYum247
    • one year ago
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    @dan815

  3. dan815
    • one year ago
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    can i suggest one thing

  4. dan815
    • one year ago
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    it might look like its less work with the formulas and stuff, but there is a simpler way to think about these questions, even though the work might seem like its a bit more

  5. YumYum247
    • one year ago
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    i don't care, spit it out!!!!

  6. dan815
    • one year ago
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    okay look

  7. YumYum247
    • one year ago
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    i am.....^_^

  8. dan815
    • one year ago
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    |dw:1439951841864:dw|

  9. YumYum247
    • one year ago
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    ive already done that....how do i figureout the horizontal displacement of the projectile.....??? :)

  10. dan815
    • one year ago
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    now u know the horizontal speed Vx, so what we need to do is find the time it is in the air for, and multiply by that time

  11. YumYum247
    • one year ago
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    yes those are the initial vertical velocity and initial horizontal velocity.....

  12. dan815
    • one year ago
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    and to find how long its in Air for, we know the Vertival speed and we have the accerelation

  13. dan815
    • one year ago
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    one thing about the launch is that the starting and ending speed for the vertical is the same

  14. dan815
    • one year ago
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    jstu the direction is opposite

  15. YumYum247
    • one year ago
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    yup i know that.... :)

  16. dan815
    • one year ago
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    so what we need to do is see the time it takes for the velicity to go to 0 and mulitply by 2

  17. dan815
    • one year ago
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    our acc is 9.8 m/s/s so Vy= 35*sin 40 lets see how long that takes to go to 0 34*sin 40 - 9.8*t = 0 every second we are going to lose 9.8 see how many secs that will take find that time multiply by 2

  18. dan815
    • one year ago
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    35*sin 40 - 9.8*t = 0 t=35sin(40)/9.8 it will be up in air twice that time so our horizontal distanve must be ? Vx*2t = 35*cos(40) * 2*35sin(40)/9.8

  19. dan815
    • one year ago
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    comesout to 123.101

  20. YumYum247
    • one year ago
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    buddy that's not even closr to the real answer!!!

  21. dan815
    • one year ago
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    maybe some error somewhre xD

  22. YumYum247
    • one year ago
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    the answer is 40m.....

  23. YumYum247
    • one year ago
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    we're looking for the horizontal displacement of the ball....

  24. YumYum247
    • one year ago
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    so we figure out the time and plug that into the horizontal equation.... D = t/v or something like that....

  25. YumYum247
    • one year ago
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    here is the solution......wait

  26. YumYum247
    • one year ago
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    1 Attachment
  27. YumYum247
    • one year ago
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    actually this topic is slightly different, its about the chaning velocity and the angle....and to find the horzontal displacement we use the formula Dh = Vi2 X sin2O/a but i wanna know what is the Vi means here....

  28. YumYum247
    • one year ago
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    is it the initial horizontal velocity or the initial vertical velocity??????

  29. dan815
    • one year ago
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    dont use formulas just learn it

  30. YumYum247
    • one year ago
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    because i'm pluging in the number and both of my x and y comps are different :( cuz when i ply them into the equation i get a differentt answer :(

  31. dan815
    • one year ago
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    i think the answer there is wrong

  32. dan815
    • one year ago
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    it should be 123m ish

  33. YumYum247
    • one year ago
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    i wanna know what is the Vi2 means here????

  34. YumYum247
    • one year ago
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    vertical or horizontal......

  35. dan815
    • one year ago
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    what are u referring to again

  36. YumYum247
    • one year ago
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    hold up.....le me draw for you :)

  37. dan815
    • one year ago
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    wait i found the mistake they made

  38. dan815
    • one year ago
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    i think they plugged in wrong numbers or something because they ahve wrong horizontal and veritcal components

  39. YumYum247
    • one year ago
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    |dw:1439952832199:dw|

  40. dan815
    • one year ago
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    yes that is the right one

  41. YumYum247
    • one year ago
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    those are my numbers...when i plug them into their equation, i don't get the right answer????

  42. dan815
    • one year ago
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    did u get the method i was showing u up there?

  43. YumYum247
    • one year ago
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    i've tried both but i ain't getting 40m :"(

  44. dan815
    • one year ago
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    40 is wrong answer

  45. YumYum247
    • one year ago
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    yuss i already know that method.....

  46. dan815
    • one year ago
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    123 is right

  47. YumYum247
    • one year ago
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    you sure bra???

  48. dan815
    • one year ago
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    yes

  49. dan815
    • one year ago
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    i did it twice so it has to be right

  50. YumYum247
    • one year ago
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    well what do you get when you pplug in your findings into the formula, do you get 40?

  51. YumYum247
    • one year ago
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    let's try

  52. dan815
    • one year ago
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    i dont use formulas i dont know, id have to derive it

  53. dan815
    • one year ago
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    what formula do u want me to derive?

  54. dan815
    • one year ago
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    u want a formula with velocity and distance only?

  55. dan815
    • one year ago
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    :)

  56. YumYum247
    • one year ago
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    change in horzontal displacement = Vi^2 X Sin2 Theta/a

  57. dan815
    • one year ago
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    okay

  58. dan815
    • one year ago
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    |dw:1439953379986:dw|

  59. YumYum247
    • one year ago
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    normally i label my vertical values and horizontal values.... Vertical: Horizontal: a = -9.8m/sec2 dh = ? Dh = 0m/s Vh = 26.81m/sec Vi = 22.49m/s up t = ? Vf = 35m/sec t = ?

  60. dan815
    • one year ago
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    okay that is fine

  61. dan815
    • one year ago
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    that is good

  62. dan815
    • one year ago
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    now look maybe u can help work this out

  63. dan815
    • one year ago
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    wait why is there a Vf=35 that doesnt eexist

  64. dan815
    • one year ago
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    Vertical Vi=22.49m/s up Vf=22.49m/s down

  65. dan815
    • one year ago
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    now tell me how much time that will take with 9.8m/s^2 acc

  66. YumYum247
    • one year ago
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    i just need to find time, so for that i can use the following dy = Vi X t + 1/2(aXt^2) i'll get a qudratic expression and fro mthere i get my time....i ply that time into the horizontal equation = V = d/t => "t = d/V"

  67. dan815
    • one year ago
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    can u just answer this simpler qusetion first please

  68. dan815
    • one year ago
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    Vertical Vi=22.49m/s up Vf=22.49m/s down how much time will this take? from this Vi to Vf?

  69. dan815
    • one year ago
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    remember acceleration =V/t right

  70. dan815
    • one year ago
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    so what is the time?

  71. YumYum247
    • one year ago
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    a = V/t yes!! :D

  72. dan815
    • one year ago
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    so how much time will it take "Vertical Vi=22.49m/s up Vf=22.49m/s down how much time will this take? from this Vi to Vf? "

  73. YumYum247
    • one year ago
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    1 sec?

  74. dan815
    • one year ago
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    no

  75. dan815
    • one year ago
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    okay how much total change was there in the velocity

  76. dan815
    • one year ago
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    |dw:1439953983112:dw|

  77. YumYum247
    • one year ago
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    wait we have 2 velocities here Vi and Vf so which one am i going to use for that....??

  78. dan815
    • one year ago
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    |dw:1439953996826:dw|

  79. dan815
    • one year ago
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    accerelation = change in velocity / change in time

  80. YumYum247
    • one year ago
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    OMG i got it..... LOL

  81. dan815
    • one year ago
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    so how much did the time change by?

  82. YumYum247
    • one year ago
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    |dw:1439954092459:dw|

  83. dan815
    • one year ago
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    okay good

  84. dan815
    • one year ago
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    so this means it took 4.5 seconds for it go up and down

  85. YumYum247
    • one year ago
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    d= t/v

  86. dan815
    • one year ago
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    in this same time it was constantly moving horizontally at some speed

  87. dan815
    • one year ago
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    Horizontal speed * ~4.5 = distance in the horizontal v*t=d

  88. YumYum247
    • one year ago
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    pellet...-_-

  89. dan815
    • one year ago
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    its oka if the answer does not match the one in the book because that one is wrong, i can show u with a simple thought experiment

  90. YumYum247
    • one year ago
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    so 4.5s X 26.81m/s

  91. dan815
    • one year ago
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    yeah

  92. YumYum247
    • one year ago
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    121 LOL Yus!!! :"D

  93. dan815
    • one year ago
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    go online and look at a football being kicked in the air, see the approximate time

  94. dan815
    • one year ago
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    it just cannot be 40m meters that is too close

  95. dan815
    • one year ago
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    they will fire him if tahts the best he can kick it

  96. dan815
    • one year ago
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    this is an opening kick off

  97. YumYum247
    • one year ago
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    no but the thing is that, this chapter covers chaning angle and velocity...so to find the hori\ontal displacement .....they gave us the formula for that.....

  98. YumYum247
    • one year ago
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    i was just following the steps....

  99. dan815
    • one year ago
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    i see i see i get it

  100. dan815
    • one year ago
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    i wouldnt worry about all that stuff trust me, you should check out the MIt physics lectures they teach kinematics properly

  101. dan815
    • one year ago
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    There is only 2 real formulas u need to know for this whole course

  102. dan815
    • one year ago
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    d(t)= displacement as a function of time d'(t)= velocity, first derivative of displacement is velocity d''(t)=acceleration, 2nd der of displament is acc from these you can get all of kinematics equations

  103. YumYum247
    • one year ago
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    so what do i do in the exams cuz this chapter and the previous chapter literally had the same types of questions....where you have the final velocity, angle and maybe one or two thingsg from the horizontal.....but virtually exactly the same types of question.....

  104. dan815
    • one year ago
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    dont worry, for all those u just need some geometry and these principles

  105. dan815
    • one year ago
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    try some more questions, post it on here if u are stuck

  106. YumYum247
    • one year ago
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    cuz i picked a question from my previous chapter on "upward movign projectiles" and applied the formula from this chapter" Changing launch angle and initial velocity....the answer i got was totally different,......

  107. YumYum247
    • one year ago
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    ok thanks ....:")

  108. YumYum247
    • one year ago
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    stay cute!!! :)

  109. dan815
    • one year ago
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    wait u got 50 something right

  110. YumYum247
    • one year ago
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    yah.....

  111. dan815
    • one year ago
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    what u got there is the change in vertical distance

  112. YumYum247
    • one year ago
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    nononoono

  113. YumYum247
    • one year ago
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    sorry different question.....

  114. YumYum247
    • one year ago
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    i'll post it if i have the energy....

  115. dan815
    • one year ago
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    okk get to work!

  116. YumYum247
    • one year ago
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    eye eye captain....:"D

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