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Alaskan Rescue Plane

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data

    An Alaskan rescue plane traveling 40 m/s drops a package of emergency rations from a height of 113 m to a stranded party of explorers.

    G = 9.8 m/s^2

    Where does the package strike the ground relative to the point directly below where is was released? Answer in units of m.

    What is the horizontal component of the velocity just before it hits? Answer in units of m/s.

    What is the vertical component of the velocity just before it hits? (Choose upwards as the positive vertical direction) Answer in units of m/s.

    2. Relevant equations

    Square root of 2h/g = time of free fall

    y = Voy(t) - 1/2gt^2

    x = Vox(t) + Xo

    3. The attempt at a solution

    I solved the first part using 40 m/s x (square root of 2h/g) = 192.09 m

    The answer to the second part was 40 because that is the constant horizontal component.

    For the third part, I tried solving for T using x = Vox(t) + Xo and plugging that into y = Voy(t) - 1/2gt^2 but could not come up with the correct answer. I believe I need to include the time it takes to hit the ground somehow but do not know.

    Any help is appreciated.

    Thanks,
    Ben
     
  2. jcsd
  3. Sep 6, 2009 #2
    part 1 & 2 look okay to me

    for part 3)
    hm for a moment ignore the horizontal component, only the vertical, well you know how far it was dropped from - displacement - 113m, you know it's acceleration - g, and you know its initial speed when it was dropped from the plane - 0, because it was still, now use your equation to find the time taken to reach the ground;

    y = Voy(t) - 1/2gt^2
     
  4. Sep 6, 2009 #3

    kuruman

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    Use

    vy=v0y - g t

    You know everything except vy.
     
  5. Sep 6, 2009 #4
    I tried this:

    Vy = Voy - gt
    Vy = 47.06 - (-9.8 x 4.80)
    Vy = 94.1 which was an incorrect answer

    If I use a positive 9.8 value, I get .02 which seems awfully low.
     
  6. Sep 6, 2009 #5

    mgb_phys

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    What is the initial vertical velocity of the dropped package ?
     
  7. Sep 6, 2009 #6
    47.06

    I used y = Voy(t) - 1/2gt^2

    113 = Voy(4.80) - 112.896
    225.896 = Voy(4.80)
    Voy = 47.06

    I thought this was the correct answer because the vertical component of the velocity should be the initial velocity in the y-direction, no?
     
  8. Sep 6, 2009 #7

    mgb_phys

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    If you drop something isn't the initial vertical velocity zero ?
     
  9. Sep 6, 2009 #8
    Yes, but the vertical component of the velocity would change and would not be zero just before it hits.

    Should I alter my y-component equation to something like:

    Voy = -1/2gt^2
     
  10. Sep 6, 2009 #9

    mgb_phys

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    In your equation Voy is the initial vertical velocity.
    If you drop something the initial velocity must add to the acceleration, you have the opposite sign as if velocity was in the opposite direction to the acceleration
     
  11. Sep 6, 2009 #10
    So going back to my first equation:

    y = Voy(t) - 1/2gt^2

    Am I supposed to use -9.8 for my g value or positive 9.8. If I use - 9.8, the math works out like this:

    y = Voy(t) - 1/2gt^2
    113 = Voy(4.80) + 112.896
    .104 = Voy(4.80)
    Voy = .0216

    To me this makes more sense because I would think that a package should arrive slowly so that the contents aren't ruined. A package traveling at 47.06 m/s would be destroyed upon impact.
     
  12. Sep 6, 2009 #11

    kuruman

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    How did the negative sign in front of (1/2)g t2 become positive when you put in the numbers? You do not seem to understand the meaning of

    vy=v0y - g t

    It says that for every second that goes by, you add -9.8 m/s to the vertical component of the velocity. The negative sign means in the "down" direction. If the package is in the air for 4.8 s you need to add 9.8*4.8 m/s in the down direction to the 0 m/s that is the initial vertical velocity. Yes. that makes a lot of meters per second, but that's why people use parachutes to drop packages (or themselves) from planes.
     
  13. Sep 6, 2009 #12
    Truthfully I always seemed to get confused on when to use a negative sign in front of 9.8. The negative sign became positive because it was minus negative 112.896. I thought that a double negative equals a positive. Or was I supposed to use a positive 9.8 value?

    Better explanation on what I did:

    -1/2gt^2

    (-1/2)(-9.8)(4.80^2) = 112.896
     
  14. Sep 6, 2009 #13
    [tex]\Delta y = v_ot - \frac{1}{2}gt^2[/tex]
    [tex]\Delta y = v_ot + \frac{1}{2}gt^2[/tex]
    You can write the equation either way. It's all about sign convention. Generally when you write the first form you take the negative sign from g, that way you know that the object would be falling.
    How you determine the sign of g (or any a). First you have to determine a sign convention, that is which way is positive velocity and which way is negative velocity. For example, dropping a ball from a height. You can make g negative and the h negative, or you can just make them both positive so you don't have to deal with negatives(you just have to realize it's really falling down). If you throw a ball downward from a height you can make all 3 negative or positive, again it's easier with positives. If you throw a ball up then it's initial velocity will be positive and g will be negative.
    If acceleration is in the same direction as motion, then it should have the same sign as the velocity. If acceleration is in the opposite direction of motion, it should have the opposite sign of velocity.
     
  15. Sep 6, 2009 #14
    Ahh, okay. So it's similar to drawing a picture where you want to set your point of reference. Thanks for the explanation. As for solving the problem:

    Vy = V0y - g t

    Vy = 0 - (9.8 x 4.8) = -47.04 which was correct.

    My last question regarding this topic is how did you derive that equation or is it just a handy one to keep around?
     
  16. Sep 7, 2009 #15
    Derived what equation?
     
  17. Sep 7, 2009 #16

    kuruman

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    The derivation is simple. "Acceleration" means "change of velocity with respect to time." When the acceleration is constant the rate of change of velocity is constant. What does that mean? That in equal time intervals, the velocity changes by equal amounts. So if the acceleration is -10 m/s/s and the velocity is 40 m/s at time zero, then it will be

    40 - 10*1 m/s at time t = 1s
    40 - 10*2 m/s at time t = 2s
    40 - 10*3 m/s at time t = 3s
    etc.

    This works equally well with any initial velocity v0y, not just 40 m/s, and any acceleration a not just -10 m/s/s. So in general

    vy = v0y + a*t

    Note that in this case we have defined "up" as positive and "down" as negative. Therefore, a = - 9.8 m/s/s, where the negative sign indicates that the acceleration is down. The symbol "g" stands for the magnitude of the acceleration of gravity, namely the number +9.8 m/s/s. I used m/s/s instead of m/s2 to stress that the acceleration is rate of change of velocity with respect to time.
     
  18. Sep 7, 2009 #17
    Thanks for the explanation.
     
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