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YumYum247

  • one year ago

Help ME Please!!!!!!!! :)

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  1. YumYum247
    • one year ago
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    @PlasmaFuzer

  2. YumYum247
    • one year ago
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    Hi there please help me!!!

  3. anonymous
    • one year ago
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    Hello give me one second..

  4. YumYum247
    • one year ago
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    ok take your time!!! :)

  5. anonymous
    • one year ago
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    ok lemme see

  6. anonymous
    • one year ago
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    Well for starters this is a matter of forces on objects.... let me ask you a question.... while the package is in the plane what forces are acting on it?

  7. YumYum247
    • one year ago
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    Potential force O_o

  8. anonymous
    • one year ago
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    ok what else?

  9. anonymous
    • one year ago
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    so the potential ENERGY the object has is given by what force? lets start there

  10. YumYum247
    • one year ago
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    the package is travelling with the velocity of the plane so whatever the seed of the plane is, that's the speed of the package inside the plane and since it's moving in a direction it'll land at an angle with respect to it's reference point( plane)

  11. anonymous
    • one year ago
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    yes yes exactley

  12. anonymous
    • one year ago
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    so the FORCES on the object in the plane are ?

  13. YumYum247
    • one year ago
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    because as the package is dropped from the plane, it'll travel both in the vertical and horizontal axis....

  14. anonymous
    • one year ago
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    dont overthink it come on

  15. anonymous
    • one year ago
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    yes yes but the forces?

  16. YumYum247
    • one year ago
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    |dw:1441073471996:dw|

  17. YumYum247
    • one year ago
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    potential and kinectic O_o

  18. anonymous
    • one year ago
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    those are the energys.... ok one of the FORCES is gravity right? gravity is always acting on everythin.... unless we flying in space right?

  19. YumYum247
    • one year ago
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    when the package was i nthe plane, the package was also travelling with the plane, but when the package was dropped, it turned into kinectic.

  20. anonymous
    • one year ago
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    and as you said before we have two axes in which we are travelling vertical and horizontal

  21. YumYum247
    • one year ago
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    Oh.....so that's what you meant so -9.81m/sec^2

  22. anonymous
    • one year ago
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    so gravity is the only force in the vertical direction... yes exactley thats g the ACCELERATION due to gravity

  23. anonymous
    • one year ago
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    what about the force in the horizontal direction? you already mentioned it was on a plane right?

  24. YumYum247
    • one year ago
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    but there is the potential force of the plane dragging it on the horizontal axis to an angle.

  25. YumYum247
    • one year ago
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    i'm referring to the package tho.

  26. YumYum247
    • one year ago
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    so what's next LOL

  27. YumYum247
    • one year ago
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    i think we first need to calculate how far on the horizontal axis the medical supplies are dropped?

  28. YumYum247
    • one year ago
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    from the campsite.....

  29. anonymous
    • one year ago
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    perhaps but I am trying to walk you through how you should think about these things in general

  30. YumYum247
    • one year ago
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    |dw:1441073888474:dw|

  31. YumYum247
    • one year ago
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    ok :)

  32. anonymous
    • one year ago
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    Yes how you are thinking about it geometrically is valid dont take what I am saying as you are wrong

  33. anonymous
    • one year ago
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    i want to try and walk you through a way of thinking about these things so you can tackle these types of probelms and harder

  34. anonymous
    • one year ago
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    so first step is to consider the forces on the package

  35. anonymous
    • one year ago
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    well for starters there is gravity acting down on the package through its center of mass

  36. YumYum247
    • one year ago
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    is there a step wise procedure to get at these types of problems??????

  37. anonymous
    • one year ago
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    but there is also the force of the plane's engines on the package

  38. YumYum247
    • one year ago
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    true!!

  39. anonymous
    • one year ago
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    i dont know what level you are in physics so i would say yes considering forces is always a valid starting point for these types

  40. anonymous
    • one year ago
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    yea :D that was what i was getting at

  41. anonymous
    • one year ago
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    if the plane was applying a force to the package then the package wouldnt be flying along with the plane no?

  42. anonymous
    • one year ago
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    however once the package is thrown out of the plane.... what forces act on it in a horizontal direction

  43. YumYum247
    • one year ago
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    kinetic force.

  44. YumYum247
    • one year ago
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    gravitational force!!!

  45. anonymous
    • one year ago
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    no im sorry yum kinetic is not a force

  46. anonymous
    • one year ago
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    yes gravity is still acting on it but that is in the vertical direction

  47. YumYum247
    • one year ago
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    yeah i thought about that too late.....:D

  48. anonymous
    • one year ago
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    you are confusing the two concepts of force and kinetic energy

  49. anonymous
    • one year ago
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    its alright no worries :D

  50. YumYum247
    • one year ago
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    :D sorry pal i'll try not to do it again!!

  51. anonymous
    • one year ago
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    dont worry about it... ok so think really hard what is acting on the package HORIZONTALLY once its been thrown out of the plane

  52. anonymous
    • one year ago
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    force i mean

  53. YumYum247
    • one year ago
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    gravitational force is acting on the package.

  54. anonymous
    • one year ago
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    correct.... and is there anything acting on the package horizontally?

  55. anonymous
    • one year ago
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    btw no is a valid answer...

  56. YumYum247
    • one year ago
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    would that be gravity as well. because the package is in the free fall, the only thing there is is gravity!!!

  57. YumYum247
    • one year ago
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    i think the gravity is making the package distace both vertically and horizontally.

  58. anonymous
    • one year ago
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    YES!!! EXACTLEY :D :D

  59. anonymous
    • one year ago
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    no horizontal force

  60. anonymous
    • one year ago
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    soooooo what does that mean for the horizontal veloctiy?

  61. YumYum247
    • one year ago
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    the package is covering a certain distance at a constant speed, but on the vertical axis, the package is accelerating downward as the gravitational force pull the package down.

  62. YumYum247
    • one year ago
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    so the horizontal component is constant but the vertical component is changing.

  63. anonymous
    • one year ago
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    YES AWESOME!!!

  64. anonymous
    • one year ago
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    ok so clearly he cant drop it right over the campsite because the constant veloctiy HORIZONTALLY of the package will take it a further distance horizontally

  65. anonymous
    • one year ago
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    during the time in which it is falling

  66. YumYum247
    • one year ago
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    yes yes!! :D

  67. YumYum247
    • one year ago
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    so how do we figure out how far ahead of the campsite the medical packages were dropped????

  68. YumYum247
    • one year ago
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    let me draw...

  69. anonymous
    • one year ago
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    ok brilliant we know the distance between the campsite and the plane (vertical distance)

  70. anonymous
    • one year ago
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    we know the initial VERTICAL velocity (pre fall) was zero (because the plane was neither climbing nor falling)

  71. YumYum247
    • one year ago
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    |dw:1441074773233:dw|

  72. anonymous
    • one year ago
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    and we know the ACCELERATION the package undergoes VERTICALLY while falling

  73. YumYum247
    • one year ago
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    -9.8m/sec^2

  74. anonymous
    • one year ago
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    Yes so how do we calculate the time of fall?

  75. YumYum247
    • one year ago
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    ok so to figure out the horizontal displacement, we first need to figure out the time it took to again that distance on the horizontal axis????????

  76. anonymous
    • one year ago
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    no in order to figure out the horizontal distance we first need to figure out the time of fall

  77. anonymous
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    in other words the time it takes to cover the vertical distance.... that we know

  78. anonymous
    • one year ago
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    do you know how to set that equation up?

  79. YumYum247
    • one year ago
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    let me try first....:)

  80. YumYum247
    • one year ago
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    Change in vertical displacement = initial velocity X t + 1/2(g X t^2) and solve for time....

  81. YumYum247
    • one year ago
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    do you want me to draw for you the equation??????????

  82. YumYum247
    • one year ago
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    |dw:1441075219970:dw|

  83. anonymous
    • one year ago
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    |dw:1441075283811:dw|

  84. anonymous
    • one year ago
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    so here we will take the intial position to be zero

  85. YumYum247
    • one year ago
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    |dw:1441075275924:dw|

  86. anonymous
    • one year ago
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    no no dont do that....

  87. YumYum247
    • one year ago
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    whyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyy!!!?!?!?!?!

  88. anonymous
    • one year ago
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    you are confusing the initial velocity int he x direction for an initial velocity in the y direction....hold on:

  89. YumYum247
    • one year ago
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    ok so i have to figure out the vertical component and horizontal component???

  90. YumYum247
    • one year ago
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    i mean do i have to.....???

  91. YumYum247
    • one year ago
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    but need the angle for that :(

  92. anonymous
    • one year ago
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    \[x(t)=x_{0}+v_{i_{x}} t + \frac{1}{2}a t^{2} = v_{i_{x}}\] && \[y(t)=y_{0}+v_{i_{y}} t + \frac{1}{2}a t^{2} =-\frac{1}{2}g t^{2} \]

  93. anonymous
    • one year ago
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    ok so we know y(T)=150m (the total fall height)

  94. YumYum247
    • one year ago
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    ok.

  95. anonymous
    • one year ago
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    so that gives us: \[y(T)=150m=\frac{-1}{2}gT^{2} \ \ \ \ \rightarrow \ \ \ \ T^{2}=-\frac{2}{g}*(150m) \approx 30.6\] \[Therefore \ \ T=\sqrt{30.6} \approx 5.52s\]

  96. anonymous
    • one year ago
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    How did I know the units were seconds.... welll look at the units on the first equation: \[T^{2}=\frac{1}{\frac{m}{s^{2}}}*m=s^{2} \ \ \ \rightarrow \ \ \ T=s\]

  97. YumYum247
    • one year ago
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    wait let me label all the given values.... Vertical: Horizontal: g = -9.8m/sec dh =? dv = 150m vh =? vi = 40m/sec t = ? vf = 0m/sec t = ?

  98. anonymous
    • one year ago
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    its a good exercise in unit analysis to check for yourself

  99. anonymous
    • one year ago
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    yes very good idea at the begining of every problem.... write down your knowns and your unknowns

  100. YumYum247
    • one year ago
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    but idk why you applied that formula tho :O

  101. YumYum247
    • one year ago
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    wait let me try doing it my wait and see if i get 5.52sec. Hold up, din't go anywhere!!! :)

  102. anonymous
    • one year ago
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    which formula? those are the kinetic motion formulas

  103. YumYum247
    • one year ago
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    but i'm not introduced to them.....formulas!!!

  104. anonymous
    • one year ago
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    i.e. the formulas to find the distance travelled as a function of time

  105. YumYum247
    • one year ago
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    to find the time only.....

  106. anonymous
    • one year ago
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    yes this is a multi step problem and my way of solving it was to find the time it falls.... multiply by the horizontal velocity and that gives you the horizontal distance travelled

  107. YumYum247
    • one year ago
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    so if i apply the way i do the time problems i should be able to get 5.52sec right?

  108. YumYum247
    • one year ago
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    give me a sec.....

  109. YumYum247
    • one year ago
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    never mind i get 140 sec or 2.3 mintues....

  110. YumYum247
    • one year ago
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    5.52 seems less for some reason.

  111. anonymous
    • one year ago
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    yea well reason this out my acceleration is approximately 10 meter per second per second and we are only travcelling 150m

  112. anonymous
    • one year ago
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    soooo if the fall time is 140 seconds.... clearly that would be wayyyyyyyyyyyyyyyyy more than 150m

  113. YumYum247
    • one year ago
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    ok..... can you please write out the equation you used to find the time..... PLease

  114. anonymous
    • one year ago
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    look above.... i already listed them

  115. YumYum247
    • one year ago
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    so....x(t)=final velocity + initail velocit + 1/2(gxt^2)

  116. anonymous
    • one year ago
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    did you get the equations i actaully worked out??? when i solved for T?

  117. YumYum247
    • one year ago
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    i can solve the equation...no problem....but help me label this.... our initial velocity is 40m/sec, our final is 0m/sec?

  118. anonymous
    • one year ago
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    and I found it to be T=5.53 seconds

  119. YumYum247
    • one year ago
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    are my initial and final ok????

  120. YumYum247
    • one year ago
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    Vi and vf?

  121. anonymous
    • one year ago
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    no... the INITIAL x velocity is 40 m/s and y is 0m/s... the FINAL x velocity is still 40 m/s and the y velocity you can solve for if you like but I didnt

  122. YumYum247
    • one year ago
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    ok...

  123. YumYum247
    • one year ago
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    and how far ahead of the campsite should the medical supplies be dropped?

  124. YumYum247
    • one year ago
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    |dw:1441077238617:dw|

  125. anonymous
    • one year ago
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    so if you take that time value I solved for.... and simply multiply by the x velocity according to the equation given above: \[x(T)=v_{i_{x}}T=(40 \frac{m}{s})*5.53s=221m\]

  126. YumYum247
    • one year ago
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    ok Thank you very very much, and god bless you!!!

  127. anonymous
    • one year ago
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    ok wait wait yum i really want to know that this made sense... could you please explain to me what we just did plz :)

  128. YumYum247
    • one year ago
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    ok so gosh....LOL ok :D

  129. YumYum247
    • one year ago
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    the only thing i didn't understand was wy idn't we use the following equation to solve for time....dv = vi X t + 1/2(a X t^2)

  130. anonymous
    • one year ago
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    because what does dv represent here? pretend like you are explaining how to solve the problem to me

  131. YumYum247
    • one year ago
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    dv is the vertical displacement from the ground up...150m at which the plane is flying at.

  132. anonymous
    • one year ago
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    again pretend like i dont understand how to solve this problem and walk me through it please :)

  133. YumYum247
    • one year ago
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    ok we have the following variables already given to us, the initial velocity of the plane = 40m/sec, dv = 150, g = -9.81m/sec^2 and the final velocity = 0m/sec

  134. YumYum247
    • one year ago
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    actually the question didn't mention the final velocity so take that out. Now the only thing that is missing here is the time...and to solve for the time the only equations that i know of are the following..

  135. YumYum247
    • one year ago
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    give me a sec please......

  136. YumYum247
    • one year ago
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    yes ok ....i got it..wait

  137. YumYum247
    • one year ago
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    change in displacement = initial velocity X t + 1/2 a X t^2

  138. YumYum247
    • one year ago
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    this is probably the only equation i can think of to solve for the time....

  139. YumYum247
    • one year ago
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  140. anonymous
    • one year ago
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    yes that is a nice list of formulas but explain to me what we did please

  141. YumYum247
    • one year ago
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    ok so we're going to find the time in which the package covered both vertical and horizontal distance....

  142. anonymous
    • one year ago
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    ok

  143. YumYum247
    • one year ago
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    but since we don't know the vertical distance from the plane to the ground, but since we have the initial velocity, gravitational force and vertical displacement...

  144. anonymous
    • one year ago
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    you mean since we dont know the horizontal distance we cover while the package is falling.... other than that good keep going

  145. YumYum247
    • one year ago
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    we can figure out with help of the constant gravity how long it'll take the package to touch down to the ground.

  146. YumYum247
    • one year ago
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    so there is this plane flying forward with the speed of 40m/sec, then there is the gravitation force being applied on the package....-9.81m/sec^2....

  147. anonymous
    • one year ago
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    ok keep doing

  148. YumYum247
    • one year ago
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    what do we do now....O_o

  149. anonymous
    • one year ago
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    ok so what I was looking for was close

  150. anonymous
    • one year ago
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    we know that gravitational acceleration.... initial velocity in both the x and y directions.... and we define the initial positions to be zero

  151. anonymous
    • one year ago
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    also we know the altitude of the plane

  152. YumYum247
    • one year ago
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    oh i forgot to mention one more thing....let me finish

  153. anonymous
    • one year ago
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    ok so the logic is that while this thing is falling... no forces act on it in a horizontal direction... meaning the x velocity remains constant during the fall

  154. anonymous
    • one year ago
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    the y velocity however keeps increasing because of the force of gravity (i.e. the presence of a gravitational acceleration g)

  155. YumYum247
    • one year ago
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    since we know our initial velocity and we know that the final will be zero m/sec, and we also know that the gravitational force is -9.81m/sec^2 is being applied on the object....we need to find what time it takes to get from that 40m/sec down to 0m/sec.

  156. anonymous
    • one year ago
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    so we know that with all the givens we can find the time it takes to fall (it is the only unknown in the equation listed above for y(t)

  157. anonymous
    • one year ago
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    we then use that time to plug into the x equation so that this equation now only has one unknown

  158. anonymous
    • one year ago
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    yum where are you getting the final velocity from?

  159. YumYum247
    • one year ago
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    Vertical: Horizontal: g = -9.8m/sec dh =? dv = 150m vh =? vi = 40m/sec t = ? vf = 0m/sec t = ?

  160. anonymous
    • one year ago
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    how did you find that though?

  161. YumYum247
    • one year ago
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    well i assume that once the package hit the ground, it's velocity will be 0m/sec

  162. YumYum247
    • one year ago
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    the package was dropped from the plane at 40m/sec. but it hits the ground, it'll be 0m/sec.

  163. anonymous
    • one year ago
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    yea in reality it goes to 0 m/s but what I am getting at is where in the formulas did the final velocity come in to play?

  164. YumYum247
    • one year ago
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    oh that's the format i use when i solve these kinds of question.......it doesn't mean every single variable has to go into the formula.

  165. anonymous
    • one year ago
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    yea exactly.... writing down everything you know is very important but one thing that comes with experience is knowing what to use and what is irrelevant

  166. YumYum247
    • one year ago
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    yah your right :(

  167. anonymous
    • one year ago
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    now there are other ways to solve this problem that perhaps use that information, but this is the most straightforward way using the equations of motion

  168. YumYum247
    • one year ago
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    ok :D

  169. YumYum247
    • one year ago
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    can you tell me one other way please....the easiest way :(

  170. anonymous
    • one year ago
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    that is the easiest way yum... the only "easier" way is to algebraically solve one equation and subsitute it into the other.... but you are still using the same two equations and doing the same procedure.... only difference is you didnt solve for the actual time T in between

  171. YumYum247
    • one year ago
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    ok. Thanks and Thanks!!! :)

  172. anonymous
    • one year ago
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    no problem yum just go over this logic again... set aside the answer and work it out... it may seem like a trivial exercise but its problems like these that allow you to make mistakes early on so later when problems are harder you dont struggle with them

  173. YumYum247
    • one year ago
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    yes i agree!!! :)

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