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YumYum247

  • one year ago

elp me please!!!!!

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  1. YumYum247
    • one year ago
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    |dw:1441153687515:dw|

  2. YumYum247
    • one year ago
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    @IrishBoy123 Police help me!!! :)

  3. YumYum247
    • one year ago
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    @dan815

  4. YumYum247
    • one year ago
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    @dan815 Dan my love come rescue me from this terrible nightmare HUN!!!! :"D

  5. YumYum247
    • one year ago
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    Any idea?????????????????//

  6. dan815
    • one year ago
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    not sure, the question is a little too vague for me

  7. dan815
    • one year ago
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    does it mean launched by compression of spring or was it just thrown up from equilibriium position

  8. dan815
    • one year ago
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    the spring i take it is not attached to the ground

  9. YumYum247
    • one year ago
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    don't worry about the compression or anything, just work with the given variables...

  10. dan815
    • one year ago
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    im trying to determine what is given

  11. YumYum247
    • one year ago
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    the g = -9.81m/sec^2, Vi = 2.3m/sec, angle = 78deg

  12. dan815
    • one year ago
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    thats just a projectile problem it has nothing to with spring

  13. YumYum247
    • one year ago
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    but we have to solve the problem in projectile sense of view....i think.

  14. dan815
    • one year ago
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    oh theres nothing to that then just find V_y

  15. dan815
    • one year ago
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    Vy= sin78* 2.3 m/s a_y= -9.8m/s^2 h=? solve

  16. dan815
    • one year ago
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    |dw:1441158571844:dw|

  17. YumYum247
    • one year ago
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    and solve for what????

  18. dan815
    • one year ago
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    height, make sure u understand what is going on

  19. dan815
    • one year ago
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    it should be natural once u understand what is going on

  20. dan815
    • one year ago
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    you have an inital velocity, and ur acceleration is tellng u how much of that velocity u are losing per second

  21. dan815
    • one year ago
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    when the object is at its peak height, its veloicty is 0, so we want to see the time it has been travelling under that acceleration for

  22. YumYum247
    • one year ago
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    ok but i can't see the picture you just posted properly, some of the content is blocked/missing..... :(

  23. dan815
    • one year ago
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    once u get the time its been travelling, you can solve for the actualy distance it travelled in that time

  24. dan815
    • one year ago
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    zoom out to see everything

  25. dan815
    • one year ago
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    http://prntscr.com/8bltc5

  26. YumYum247
    • one year ago
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    ok...bit one more thing, there is a formula for this but i'm sure if that would work...let me post it...hold up :)

  27. YumYum247
    • one year ago
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    H=u2sin2(θ)2g

  28. YumYum247
    • one year ago
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    it looks something like this, le me draw.

  29. YumYum247
    • one year ago
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    |dw:1441159093595:dw|

  30. YumYum247
    • one year ago
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    is this right???????????

  31. dan815
    • one year ago
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    im thinking that is from equating the kinetic and potential energy together

  32. dan815
    • one year ago
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    see if both ways gives the same answer if it does then its right

  33. YumYum247
    • one year ago
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    but what if it doesn't???:O

  34. dan815
    • one year ago
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    1/2 m * V_Y^2 = mgh therefore V*sin ^2 (78)/2g = h

  35. dan815
    • one year ago
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    i think you are relying on formulas too much, you have to really understand physics or it wont work

  36. YumYum247
    • one year ago
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    ok what to do for now???? which one tho????

  37. dan815
    • one year ago
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    check both

  38. YumYum247
    • one year ago
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    aight.....!!! Thanks!!! :"D

  39. dan815
    • one year ago
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    i know whats gonna happen both ways, but i want you to explore

  40. YumYum247
    • one year ago
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    ok dear, i'll try to explore. Thank!!

  41. YumYum247
    • one year ago
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    what is the formula for projectile motion, i didn't get it for a sec.... t = Vf +Vi + 1/2(G+t^2)

  42. YumYum247
    • one year ago
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    @dan815

  43. dan815
    • one year ago
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    |dw:1441161107807:dw|

  44. dan815
    • one year ago
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    |dw:1441161320080:dw|

  45. dan815
    • one year ago
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    u can do this integral too to see if gives u the same distance

  46. YumYum247
    • one year ago
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    nooooooooooooooooooooooooooooooooooooooi 2 is not the power.....

  47. dan815
    • one year ago
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    http://www.wolframalpha.com/input/?i=h%3D2.3^2*sin^2%2878%29%2F%282*9.8%29

  48. YumYum247
    • one year ago
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    2 multiplies with the sign...i have the formula sheet with me rn.

  49. dan815
    • one year ago
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    both ways same answer

  50. dan815
    • one year ago
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    the formula is wrong

  51. dan815
    • one year ago
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    i showed u how the formula is derived

  52. dan815
    • one year ago
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    it comes from kinetic1+potenrial1= kinetic2+ potential2

  53. YumYum247
    • one year ago
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    ok......

  54. dan815
    • one year ago
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    so in the beginning theres only kinetic as the height is 0 and in the end theres only potential as the velocity in y direction is 0

  55. dan815
    • one year ago
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    1/2mv^2=mgh m cancel both sides 1/2 V^2=gh so h=1/2v^2/g

  56. YumYum247
    • one year ago
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    yah makes sense......

  57. dan815
    • one year ago
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    or h=v^2/2g

  58. dan815
    • one year ago
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    now v^2 is the V_y^2so

  59. YumYum247
    • one year ago
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    wait what do you mean by Vy tho...vertical velocity O_o

  60. dan815
    • one year ago
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    V_y^2=(Vi*sin 78)^2 =Vi^2 * sin^2 (78)

  61. dan815
    • one year ago
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    yes Vy is the vertical velocity

  62. dan815
    • one year ago
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    Vi is more like the speed or the veloicty in that specific direction

  63. YumYum247
    • one year ago
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    ok thanks a lot.....i mean alot. :)

  64. dan815
    • one year ago
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    okay but u better understand this stuff

  65. dan815
    • one year ago
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    the most important thing is you are able to visualize and work through it in your head

  66. dan815
    • one year ago
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    or paper

  67. YumYum247
    • one year ago
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    yah i'm doing it....it takes time bro, gimme sumtime bro.......

  68. YumYum247
    • one year ago
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    |dw:1441162910616:dw|

  69. YumYum247
    • one year ago
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    @dan815 i'm stuck!!!! :"(

  70. dan815
    • one year ago
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    what kinda equation is that

  71. dan815
    • one year ago
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    its h(t) = 0 + v_y*t+1/2 * a*t^2

  72. dan815
    • one year ago
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    h(t) is notation for h as a function of t

  73. dan815
    • one year ago
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    Vy= vi*sin (78) t=Vy/9.8 h = 0 + v_y*t+1/2 * a*t^2

  74. YumYum247
    • one year ago
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    i didn't post the whole thing but i followed your steps.....

  75. YumYum247
    • one year ago
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    Vy= vi*sin (78) t=Vy/9.8 h = 0 + v_y*t+1/2 * a*t^2 if you simplify this you get something i posted.....

  76. dan815
    • one year ago
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    \[V_y= vi*sin (78)\\ t=V_y/9.8 \\ h= 0 + v_y*t+1/2 * a*t^2\]

  77. YumYum247
    • one year ago
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    that's a quadratics expression.

  78. YumYum247
    • one year ago
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    if you rearrange it.

  79. dan815
    • one year ago
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    its not a quadratic u know what t is already u plug that in

  80. YumYum247
    • one year ago
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    omg...i didn't see that....sory....:) let me try again....

  81. dan815
    • one year ago
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    Vi=2.3 Vy=2.3*sin(78) t=Vy/9.8 h=0+Vy*t + 1/2 a*t^2

  82. YumYum247
    • one year ago
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    don't say anything just let me take a look at this with my eyes and brain open...

  83. YumYum247
    • one year ago
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    i got it....

  84. YumYum247
    • one year ago
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    Thanks!!!

  85. dan815
    • one year ago
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    check with h=Vy^2/(2g)

  86. YumYum247
    • one year ago
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    yah that works for me....

  87. YumYum247
    • one year ago
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    h = vertical velocity/2g gives me the right answer....

  88. YumYum247
    • one year ago
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    Anyways don't stress yourself..... go watch a porn to cool off a lil.... :"D i'll get this done on my own now!! :)

  89. YumYum247
    • one year ago
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    btw the example you gave me on wolframe didn't have the time input for T.

  90. YumYum247
    • one year ago
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    |dw:1441164804108:dw|

  91. YumYum247
    • one year ago
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    @dan815 please take a look at this.

  92. YumYum247
    • one year ago
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    i have to use quadratics formula here....

  93. YumYum247
    • one year ago
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    yes i get that...but what happened at the very last step....

  94. YumYum247
    • one year ago
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    idk if the website used quadratics formula, or what to get the answe!!!

  95. YumYum247
    • one year ago
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    im stuck at the very last part....

  96. YumYum247
    • one year ago
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    |dw:1441165539159:dw|

  97. YumYum247
    • one year ago
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    never mind i got it....

  98. YumYum247
    • one year ago
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    Thanks !!

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