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Falling woman

  1. Jul 2, 2007 #1
    1. The problem statement, all variables and given/known data

    A woman is reported to have fallen about 139 ft from the 17th floor of a building, landing on a metal ventilator box that she crushed to a depth of 20 in. She suffered only minor injuries.

    (a) Ignoring air resistance, calculate the speed of the woman just before she collided with the ventilator.

    Your answer differs from the correct answer by 10% to 100%. ft/s
    (b) Calculate her average acceleration while in contact with the box.

    (c) Modeling her acceleration as constant, calculate the time it took to crush the box.



    2. Relevant equations
    [tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex]
    [tex]v = v_0 + at[/tex]


    3. The attempt at a solution

    I'm stuck on the first part. I found a previous thread that had covered some of the solution methods so here goes.

    x = 0 since we want her speed before she hit the ground and I converted x_0 to 1668inches (from 139ft). a is a constant that is -9.8m/s^2 or -32ft/s^2 (I will be using -32 since the answer calls for units in feet). That gives...
    [tex] 0 = 1688 + v_0t + \frac{1}{2}(-32)t^2[/tex]

    The other thread that I viewed mentioned setting [tex]v_0[/tex] to 0 and solving for t, although I'm not sure the purpose behind this or how would it help. What would you do after solving for t?
     
    Last edited: Jul 2, 2007
  2. jcsd
  3. Jul 2, 2007 #2

    Dick

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    You are doing great so far. But why convert feet to inches? Just work in feet. Putting v0=0 just means she started with 0 velocity. The just get t^2 alone on one side of the equation and take the square root of both sides.
     
  4. Jul 2, 2007 #3

    Doc Al

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    Staff: Mentor

    Why in the world did you convert the height to inches! You must use consistent units.

    Then you could make use of the second of your two relevant equations to solve for the speed upon impact.
     
  5. Jul 2, 2007 #4
    I think I see it now... converting those units was just a dumb mistake.
    Anyway...

    Since Vo = 0, the math became
    [tex] 0 = 139 + \frac{1}{2}(-32)t^2[/tex]
    t=the time she hit the ventilator=2.95 seconds

    The time it took can be plugged into...
    [tex]v = v_0 + at[/tex]
    again, Vo becomes 0 again since her initial velocity was zero, -32ft/s^2 for acceleration constant giving...
    [tex]v = -32(2.95)[/tex]
    or -94.4ft/s
    right?
     
    Last edited: Jul 2, 2007
  6. Jul 2, 2007 #5

    Dick

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    Quite right.
     
  7. Jul 2, 2007 #6
    Excellent, thanks for the help! The website recorded 94.4 as the correct answer.

    Now for the B Part. It asks for the woman's average acceleration while in contact with the box.

    1. The problem statement, all variables and given/known data

    I've been working through this in a similar manner using the position function. The following is a schematic of what I think the problem looks like.

    |
    |
    | 139 ft
    |
    |__
    | 20 inches (1.67 feet)

    2. Relevant equations
    [tex]a_{ave} = \Delta v / \Delta t[/tex]
    [tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex]
    [tex]v = v_0 + at[/tex] ??

    3. The attempt at a solution
    Basically in my attempt I looked for the initial and final velocities from the point of contact with the ventilator along with their times and plugged them into
    [tex]\frac{v - v_0}{t - t_0}[/tex]

    [tex]x = x_0 + v_0t + \frac{1}{2}at^2[/tex]
    I set [tex]x[/tex]=1.67ft (converted from 20 inches),
    [tex]v_0[/tex] to 94.4 (the answer to the previous question--or do we assume velocity is zero when she hits the ground again?),
    and solved for [tex]t[/tex] using the quadratic formula to get
    .0176 seconds
    [tex]1.67 = 94.4t + \frac{1}{2}(32)t^2[/tex]
    [tex]t = .0176[/tex]

    [tex]v = v_0 + at[/tex]
    I plugged .0176 into here to get the final velocity and got 94.96m/s.
    [tex]v = 94.4 + (32*.0176)[/tex]

    Finally, I plugged this into
    [tex]\frac{v - v_0}{t - t_0}[/tex] to get 32, a wrong answer :(
    Any ideas on what I did wrong? Is this problem solved in a different or similar manner?
     
    Last edited: Jul 2, 2007
  8. Jul 2, 2007 #7

    Doc Al

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    Note that the last two equations assume a constant acceleration, which will be the average acceleration during the collision. You'll need to combine both of those last two equations to solve this problem. (Or you can look for another kinematic relationship that already relates v, d, and a. Look here: Basic Equations of 1-D Kinematics)

    Your mistake is using a value of 32 f/s^2 as the acceleration. (The acceleration is what you are trying to solve for!) Note that during the collision the woman is not in free fall, so her acceleration is not simply the acceleration due to gravity.
     
  9. Jul 2, 2007 #8
    I'm beginning to see what you mean and I'm currently deriving a formula to solve the problem, can we still assume [tex]v_0[/tex] is the velocity from the answer to (a) in this case?
     
  10. Jul 2, 2007 #9

    Doc Al

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    Absolutely.
     
  11. Jul 2, 2007 #10
    I'm still having a little trouble here. Would we assume Xo and X to be 0 and 1.67 feet since the question only asks for contact with the box? And can we assume the final velocity to be 0 since she eventually stops moving after 1.67 feet?

    What I did was calculated the time it took for the woman to stop moving after hitting the ground given by
    [tex]x = x_0 + (1/2)(v_0 + v)t[/tex]

    (Here it is inserted with my values)
    [tex]1.67 = 0 + (1/2)(94.4 + 0)t[/tex]
    t=.035 seconds
    __________________________________________
    [tex]\frac{v - v_0}{t - t_0}[/tex]

    (My values here)
    [tex]\frac{0 - 94.4}{.035 - 0}[/tex]
    =-2697.14m/s^2
    .... doesn't seem right to me though.
     
  12. Jul 2, 2007 #11

    Dick

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    Don't let this go to your head, but you are actually quite good at these problems (though you are doing some sign arithmetic in your head). I get about the same answer. It is nearly 100g's of acceleration. We can assume the survivor was very lucky.
     
    Last edited: Jul 2, 2007
  13. Jul 2, 2007 #12
    WOOT. Got it right! I just hate the website I use for homework since it only allows us 3 tries per problem so I was freakin out. Thanks for the nod of confidence!
     
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