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A Person in Free Fall Newton's 2nd Law

  1. Oct 15, 2012 #1
    A Person in "Free Fall" Newton's 2nd Law

    1. The problem statement, all variables and given/known data
    An 80kg aviator is in free fall and acquires a velocity of 60 m/s and then opens his parachute.
    After falling an additional 30m, his velocity has been reduced to 20 m/s. What is the average acceleration of the aviator while his fall is being checked? What is the average force being exerted by the parachute?


    2. Relevant equations
    Force = Mass * acceleration
    Average acceleration = change in velocity/change in time


    3. The attempt at a solution
    Any help would be appreciated, Dont understand how to go about problem
     
  2. jcsd
  3. Oct 15, 2012 #2

    HallsofIvy

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    Re: A Person in "Free Fall" Newton's 2nd Law

    If a person has intial speed [itex]v_0[/itex] and constant acceleration [itex]a[/itex] has, after time t, speed [itex]at+ v_0[/itex] and will have gone distance [itex](a/2)t^2+ v_0t[/itex]. So we will fall "an additional 30m" when [itex](a/2)t^2+ 60t= 30[/itex]. Solve the quadratic equation [itex](a/2)t^2+ 60t- 30= 0[itex] for t. Of course, that will depend on a. Put that value of t into [itex]at+ 60= 20[/itex] and solve that equation for a.
     
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