Finding time of travel by falling object in drag coeeficient

  • Thread starter vorse
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  • #1
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Homework Statement



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?



Homework Equations



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



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?
 

Answers and Replies

  • #2
HallsofIvy
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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.
 
  • #3
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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
 
  • #4
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Bump*
Don't know the rule for bumping threads, but if this is illegal, I'll never do this again.
 

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