1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding time of travel by falling object in drag coeeficient

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data

    An object with mass 10kg is dropped from a height of 200m. Given that the constant k in
    the equation is 2.5Nsm^-1
    mv'(t) = −mg − kv(t)

    approximately how many seconds does the object hit the ground?



    2. Relevant equations

    v' + (k/m)v = -g



    3. The attempt at a solution

    p(t) = k/m h(t) = (k/m)t e^h(t) = e^(k/m)t

    e^(k/m)t v = -gm/k e^(k/m)t +C

    v(t) = -gm/k + Ce^-(k/m)t

    v(0) = 0 ; C = gm/k

    v(t) = -gm/k + gm/k e^-(k/m)t

    S(t) - S(0) = integration of v(t) = -gmt/k - gm^2/k^2 e^-kt/m ; evaluate from 0 to t

    0 = -39t - 156.8e^(-1/4) + 156.8 ; after substituting m = 10kg ; g = -9.8; and k = 2.5


    Now I must solve for t, but how can I do this? I can't apply the quadratic formula can I?
     
  2. jcsd
  3. Sep 15, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Solve 0= -39t - 156.8e^(-1/4) + 156.8 for t? That's a simple linear equation.

    39t= -156.8e^(-1/4) + 156.8 so t= (-156.8e^(-1/4) + 156.8)/39.
     
  4. Sep 15, 2009 #3
    0 = -39t - 156.8e^(-1/4)t + 156.8 ; after substituting m = 10kg ; g = -9.8; and k = 2.5

    sorry, forgot the t from this equation; if you follow the work I did, you'll see. Still, not sure how to solve for t
     
  5. Sep 16, 2009 #4
    Bump*
    Don't know the rule for bumping threads, but if this is illegal, I'll never do this again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding time of travel by falling object in drag coeeficient
  1. Fall with drag (Replies: 6)

Loading...