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Air Resistance and drag coefficien

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data

    An object of mass 10kg is projected upwards (from ground level) with initial velocity 60m/s. It hits the ground 8.4 seconds later.

    Find the drag coefficient, k.

    2. Relevant equations

    dv/dt + rv = -g, where r = k/m

    3. The attempt at a solution

    I have used the integrating factor of e^rt to give me the final equation:

    v = -g/r + C/e^rt

    I then plug in the initial values to get:

    60 = -98/k + C

    I am not sure what to do next. We are given an impact time of 8.4 seconds. Do I assume that at this instant the velocity is -60 m/s (i.e. the exact opposite of the initial)? Or can I assume that at time 4.2 seconds the velocity is equal to 0? In either case, I am not sure how to solve the equation so that I only have one variable (e.g. just k or just C).

  2. jcsd
  3. Sep 29, 2008 #2
    I don't follow your logic, why would you do this? I would start by making a FBD of the object. Your drag force will obviously be a function of velocity but you should come up with a fairly simple integral based off of the golden kinematics equations.
  4. Sep 29, 2008 #3
    Solved it.

    I had to use e^-rt, use the initial velocity to give me a value for C, substitute that back in and then integrate with respect to t to give me height. From there you know that at t=0 and t=8.4, the height is 0. You can then calculate r and because you know the mass, k.

    Use that r value for the velocity and you can then solve the velocity at t=8.4. The maximum height will be when the velocity equation is equal to 0.


    Took me 2 hours, but I worked it out.
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