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Kinematics/Projectile Motion Question?

  1. Feb 19, 2013 #1
    1. The problem statement, all variables and given/known data
    If one jump from an airplane 4572m above the ground. if the terminal speed of a human is 53m/s (without a parachute) how long would it take u to hit the ground and how fast would u be going when u hit the ground?





    3. The attempt at a solution

    I answered the second part with just 53m/s since its the terminal speed

    For part 1 however, I first found the time it takes for the human to reach 53m/s soo:
    9.81 = (53 - 0)/t....... t = 5.4s.....With that i found the distance it travels during this time period : d = (0)(5.4) + (1/2)(9.81)(5.4)^2........ d = 143m

    With this distance, I subtracted it from the original distance: 4572 - 143 = 4429m
    With that distance, I found the time it takes so: t = 4429/53 = 83.56s..... then I added the 2 times: so 83.56 + 5.4 = 88.96s = total time


    Is this correct? or is there a step missing or am I going in the wrong direction?
     
  2. jcsd
  3. Feb 19, 2013 #2

    mfb

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    A human does not accelerate with 9.81m/s^2 until it reaches its final velocity and stays at that velocity (with zero acceleration) afterwards. It should give a reasonable approximation, however. The real acceleration begins at 9.81m/s^2 and goes down as the speed gets closer and closer to the terminal speed (which is never reached in an ideal model). A better evaluation would need a differential equation (and its solution).
     
  4. Feb 19, 2013 #3
    So would I do something like this:
    d = (Vi)t +(1/2)(a)t^2
    4572m = 0m + (0m/s)(t) + (1/2)(-9.81m/s^2)(t^2)
    4572m = (-4.905m/s^2)(t^2)
    -932.1s^2 = t^2
    30.53s = t

    Vf = Vi + at
    Vf = 0m/s + (-9.81m/s^2)(30.53s)
    Vf = -299.5m/s
    But since you cannot exceed terminal you will still be going at 53m/s?
     
  5. Feb 19, 2013 #4

    mfb

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    No, that would neglect air resistance completely...
    Do you know how to calculate the force due to air drag? With F=m*a, this gives a velocity-dependent acceleration.
     
  6. Feb 19, 2013 #5
    I think we are supposed to neglect air resistance
     
  7. Feb 19, 2013 #6

    mfb

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    The terminal speed is the result of air resistance (and gravity) only.
     
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