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ChrisV Group Title

s(t)=-4.9+vt+s To estimate the height of a building, a stone is dropped from the top of the building into a pool of water at ground level. How high is the building if the splash is seen 5.6 seconds after the stone is dropped?

  • 2 years ago
  • 2 years ago

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  1. ChrisV Group Title
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    s(t)=-4.9t+vt+s

    • 2 years ago
  2. Awstinf Group Title
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    hmm are you aloud to use derivatives?

    • 2 years ago
  3. ChrisV Group Title
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    missed a t >.<

    • 2 years ago
  4. Awstinf Group Title
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    -4.9t^2 I think you mean :)

    • 2 years ago
  5. ChrisV Group Title
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    yea

    • 2 years ago
  6. ChrisV Group Title
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    -4.9t^2 +vt+s

    • 2 years ago
  7. ChrisV Group Title
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    and yes i can use derivatives

    • 2 years ago
  8. ChrisV Group Title
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    im just not sure how to use the information given

    • 2 years ago
  9. ChrisV Group Title
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    im so confused right now

    • 2 years ago
  10. ChrisV Group Title
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    this rate of change thing is going to do me in

    • 2 years ago
  11. Awstinf Group Title
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    hmm so vo and so are both unknown?

    • 2 years ago
  12. ChrisV Group Title
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    yes but i am assuming s0 would = 0 at 5.6 seconds

    • 2 years ago
  13. ChrisV Group Title
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    since thats impact

    • 2 years ago
  14. Awstinf Group Title
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    yeah thats right the vo is just the pain in the a part of the problem

    • 2 years ago
  15. Awstinf Group Title
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    wait wait wait, it seems to be lacking the classic physics wording of it was dropped with an unknown intial velocity, it being dropped and not thrown I would maybe go as far to say the the intial velocity is 0

    • 2 years ago
  16. Awstinf Group Title
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    that seems like it might be too easy though

    • 2 years ago
  17. Awstinf Group Title
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    we dont drop things with an intial velocity though lol, thats against the definition of dropping lol

    • 2 years ago
  18. ChrisV Group Title
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    I am so lost on any problems dealing with rate of change

    • 2 years ago
  19. ChrisV Group Title
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    I may have to drop this course if I do not pick these up soon :(

    • 2 years ago
  20. Awstinf Group Title
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    :(

    • 2 years ago
  21. Awstinf Group Title
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    4.9(5.6^2) should be your answer assuming that dropping means letting go from rest

    • 2 years ago
  22. ChrisV Group Title
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    I am having a hard time telling where the information they give me plugs in

    • 2 years ago
  23. ChrisV Group Title
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    Say they give me a distance of 108 meters, I assume thats s

    • 2 years ago
  24. ChrisV Group Title
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    if they give me a velocity that v0t

    • 2 years ago
  25. ChrisV Group Title
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    and a measurement of time = t

    • 2 years ago
  26. ChrisV Group Title
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    so if they give me t and vot

    • 2 years ago
  27. ChrisV Group Title
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    i find the derivative

    • 2 years ago
  28. ChrisV Group Title
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    and solve for the value of t

    • 2 years ago
  29. ChrisV Group Title
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    but when they give me t and s im lost lol

    • 2 years ago
  30. Awstinf Group Title
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    s(t)= -4.9t^2+v0t+108?

    • 2 years ago
  31. ChrisV Group Title
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    and when they only give me t i am more than lost

    • 2 years ago
  32. ChrisV Group Title
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    no that more like my next problem, but if i were given that question

    • 2 years ago
  33. Awstinf Group Title
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    Here ill read type you a problem from my book and try to help clarify things

    • 2 years ago
  34. ChrisV Group Title
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    i do not know what to do

    • 2 years ago
  35. Awstinf Group Title
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    read-type*

    • 2 years ago
  36. Awstinf Group Title
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    Hokay so, A person standing at the edge of a cliff throws a rock directly upward. it is observed that 2 seconds later the rock is at its maximum height and that 5 seconds after that, it hits the ground at the base of the cliff?

    • 2 years ago
  37. Awstinf Group Title
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    questions like that?

    • 2 years ago
  38. ChrisV Group Title
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    i have no idea how to solve that

    • 2 years ago
  39. Awstinf Group Title
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    But they are questions like that correct?

    • 2 years ago
  40. Awstinf Group Title
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    I'll show you how to appoarch it if it is

    • 2 years ago
  41. Awstinf Group Title
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    (this is straight from my change of rate section)

    • 2 years ago
  42. ChrisV Group Title
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    hmmmm on my original question

    • 2 years ago
  43. Awstinf Group Title
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    yeah?

    • 2 years ago
  44. ChrisV Group Title
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    s(t)=-4.9+vt+s v(t)=s'(t)

    • 2 years ago
  45. ChrisV Group Title
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    oops 4.9t^2

    • 2 years ago
  46. ChrisV Group Title
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    so -9.8t+v would be the derivative

    • 2 years ago
  47. Awstinf Group Title
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    yeah

    • 2 years ago
  48. ChrisV Group Title
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    -9.8(5.6)+v=0 v= 54.88

    • 2 years ago
  49. ChrisV Group Title
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    would that be the velocity?

    • 2 years ago
  50. Awstinf Group Title
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    thats your impact velocity

    • 2 years ago
  51. ChrisV Group Title
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    ok so when s=0 Velocity = 54.88?

    • 2 years ago
  52. Awstinf Group Title
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    assuming that v0=0 yes

    • 2 years ago
  53. ChrisV Group Title
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    now i confused myself lol

    • 2 years ago
  54. Awstinf Group Title
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    hahaha remember the derivative of the distance equation is the velocity at time equation

    • 2 years ago
  55. ChrisV Group Title
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    -4.9t^2 +vt+s

    • 2 years ago
  56. ChrisV Group Title
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    well if s=distance

    • 2 years ago
  57. ChrisV Group Title
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    at 5.6 secs s=0 since thats impact right?

    • 2 years ago
  58. Awstinf Group Title
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    yes

    • 2 years ago
  59. Awstinf Group Title
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    wait nno no s0 is not o

    • 2 years ago
  60. Awstinf Group Title
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    0*

    • 2 years ago
  61. Awstinf Group Title
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    here label the equation like this it makes it more clear. So 4.9t^2+v0t+s0= s

    • 2 years ago
  62. Awstinf Group Title
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    so at time t=5.6s s would be 0 not s0

    • 2 years ago
  63. Awstinf Group Title
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    other wise you would get a negative height, it really doesnt matter which way you do it but you should do it right when you are first doing it.

    • 2 years ago
  64. ChrisV Group Title
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    hehe

    • 2 years ago
  65. ChrisV Group Title
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    well i am trying to understand

    • 2 years ago
  66. Awstinf Group Title
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    I'm sorry I really havent been too much help but that question is horribly worded. If I were in your shoes I would assume that the initial velocity is 0 as progress from there. As long as you understand the relationships between the equations, it will make solving these very simple. Drawing whats going on helps

    • 2 years ago
  67. ChrisV Group Title
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    -4.9t^2+vt+s 9.8t+v is the derivative correct?

    • 2 years ago
  68. Awstinf Group Title
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    yes sir

    • 2 years ago
  69. ChrisV Group Title
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    with that i can get the velocity at impact

    • 2 years ago
  70. ChrisV Group Title
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    -9.8(5.6)+v=0 v= 54.88

    • 2 years ago
  71. Awstinf Group Title
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    yes but you need to know what v initial is

    • 2 years ago
  72. ChrisV Group Title
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    well wouldnt that be as simple as making t 0?

    • 2 years ago
  73. Awstinf Group Title
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    I dont think you can do that because you are stating that v(5.6)=0

    • 2 years ago
  74. Awstinf Group Title
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    Its hard to explain

    • 2 years ago
  75. ChrisV Group Title
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    yea i see lol

    • 2 years ago
  76. Awstinf Group Title
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    Try reposting this in the physics section, they are really good at explain this kind of stuff

    • 2 years ago
  77. Awstinf Group Title
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    http://www.blurtit.com/q3330030.html

    • 2 years ago
  78. ChrisV Group Title
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    will do thanks

    • 2 years ago
  79. Awstinf Group Title
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    I was right :)

    • 2 years ago
  80. Awstinf Group Title
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    the initial velocity is 0

    • 2 years ago
  81. ChrisV Group Title
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    :)

    • 2 years ago
  82. ChrisV Group Title
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    :)

    • 2 years ago
  83. Awstinf Group Title
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    Sorry I was trying to be ambiguous I didnt want to give you improper information

    • 2 years ago
  84. Awstinf Group Title
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    I am in your same level class lol

    • 2 years ago
  85. Awstinf Group Title
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    I just do about 6 hours of hw a day

    • 2 years ago
  86. ChrisV Group Title
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    i took too many courses

    • 2 years ago
  87. ChrisV Group Title
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    falling behind

    • 2 years ago
  88. Awstinf Group Title
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    yeah it blows I have no social life atm just study study

    • 2 years ago
  89. ChrisV Group Title
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    com 101 is more time consuming than i thought

    • 2 years ago
  90. Awstinf Group Title
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    :( yeah thats the draw back of underestimating courses. I'm taking physics, chem, physics lab, calc and I really thought this semester would be a tad less stressful

    • 2 years ago
  91. ChrisV Group Title
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    chem, com 101, calc and us history

    • 2 years ago
  92. ChrisV Group Title
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    history and chem is ok, but com 101 is taking way to much of my time

    • 2 years ago
  93. Awstinf Group Title
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    us history=gun to my head i finished all of my gen ed first so I could figure out what I wanna do

    • 2 years ago
  94. ChrisV Group Title
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    taking way to much time writing and preparing for speeches

    • 2 years ago
  95. Awstinf Group Title
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    Thats why I stick to courses that strictly deal with numbers

    • 2 years ago
  96. Awstinf Group Title
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    lol

    • 2 years ago
  97. ChrisV Group Title
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    lol its a req for my degree or i would never touch it

    • 2 years ago
  98. Awstinf Group Title
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    :( I just wanna take comparitive virology already, all of these courses building up to it are just BS

    • 2 years ago
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