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Rock free fall

  1. Jun 13, 2009 #1
    1. The problem statement, all variables and given/known data
    A stone is thrown upward from ground level. The initial speed is 32 ft/sec
    a) In how many seconds will the stone hit the ground?
    b) How high will it go?
    c) With what minimum speed should the stone be thrown so as to reach a height of 36 ft?


    2. Relevant equations

    y(t) = -16(t)2 + v0(t) + y0

    3. The attempt at a solution

    A) y(t) = -16(t)2 + 32(t) +0

    0 = -16(t)2 +32
    t = 2 sec till it hits the ground

    B)y'(t) = v(t) = -32t + 32
    0 = -32t +32
    t = 1 sec to reach max

    -16 +32 = 16 ft max

    C) y(t) = 36 ft y(t) = -16(t)2 +32(t)
    -16(t)2 +32(t) = 36
    +16(t)2 -32(t) +36 = 0
    4(4(t)2 - 8(t) + 9 = 0


    Then I assumed I was doing it wrong : /

    I continued onward to get (t-1)2 = -(5/4)
    t = 41/16
    v(41/16) = -32(41/16) +32 = -50 = 50 ft/s

    I got part c wrong naturally, any help on where it went wrong would be appreciated. Or if I did the first two wrong and just got the right answer. Either way, thanks for your time.
     
    Last edited: Jun 13, 2009
  2. jcsd
  3. Jun 13, 2009 #2

    Cyosis

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    This belongs in the introductory physics forums.

    You did A and B correctly.

    C is a bit sloppy although not necessarily the wrong approach. You calculated t incorrectly, and when you're using [itex]v=v_0-gt[/itex], you use [itex]v_0=32 \;\;ft/sec[/itex], but you're trying to solve [itex]v_0[/itex]! [itex]v_0=32 \;\;ft/sec[/itex] belongs to the previous question and of course if the initial velocity is the same as in part a then the maximum height will be the same so this cannot be true.

    The best way would be to start with the formula for the velocity, so you don't have to use the ABC formula.

    [itex]v=v_0-gt[/itex], what value is v at its maximum height?
     
  4. Jun 13, 2009 #3
    Velocity = 0 at max so the initial velocity is equivalent to gravity(t)?

    0 = V0 - 16t

    16t = V0

    or V0/16 = t
    :uhh:
     
    Last edited: Jun 13, 2009
  5. Jun 13, 2009 #4

    Cyosis

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    Since you have t now you can plug it into y(t) and solve for v0.
     
  6. Jun 13, 2009 #5
    So

    y(V0/16) = -16(V0/16)2 + V0(t)

    y(V0/16) = (V0)2/16 + V0(V0/16)

    y(V0/16) = (V0)2/16 + (V0)2/16
    y(V0/16) = 2V02/16

    36 = 2V02/16

    576 = 2V02

    288 = V02

    16.9 = V0

    :cry: There is no hope it seems.

    Step X) Answer is 48 b/c it is in the back of book, thus I win for now.
     
  7. Jun 13, 2009 #6

    Cyosis

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    Since I don't use the imperial system I missed this at first, but isn't g 32 ft/sec in the imperial system?

    Where did the minus go?
     
  8. Jun 13, 2009 #7
    g is 32 ft /s but the formula is y(t) = - (1/2)gt2 + V0 Y0

    The minus did vanish but then wouldnt I get y(t) = 0? Well 36 = 0

    -V02/16 + V02/16
     
  9. Jun 13, 2009 #8

    Cyosis

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    Yes but you didn't use that formula, you used

    to determine t.
     
  10. Jun 13, 2009 #9
    Finally, I think I love you tbh.

    V0/32 = t

    -16V02/1024 + V02/32

    -16V02/1024 + 32V02/1024

    36 = 16V02

    36864 = 16V0

    2304 = V0

    48 ft/s = V0

    Thank you for your help, I really appreciate it as I was close to burning the book.:blushing:
     
  11. Jun 13, 2009 #10

    Cyosis

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    You're welcome but you should really be less sloppy. Your answer is correct but if you wrote it down like that an exam I fear you would get very little points.

    All good so far.

    then

    You multiplied the right side by 1024, but didn't multiply the left side.

    It should be [itex]36*1024=16v_0^2[/itex]

    Here the 1024 is back so the left hand side is correct now, but now you lose a square on the right hand side.

    It should be

    [itex]36864=16v_0^2[/itex]

    You divided by 16 on both sides correctly, but again the square is gone.

    It should be [itex]2304=v_0^2[/itex]

    I know you took the square root on both sides here, but as you've written the last two equations down it reads [itex]v_0=2304ft/s=48ft/s[/itex] which obviously isn't right.

    Tip: I personally prefer to not fill in the numbers until the end. This way you're less prone to make mistakes. We know that [itex]t=\frac{v_0}{g}[/itex], therefore [itex]y_{max}=\frac{v_0^2}{g}-\frac{1}{2}\frac{v_0^2}{g}=\frac{1}{2}\frac{v_0^2}{g}[/itex]. Solve for v_0 gives [itex]\sqrt{2 g y_{max}}=v_0[/itex]. It is exactly the same method but a lot cleaner. Now enter the numbers and obtain your answer.
     
    Last edited: Jun 13, 2009
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