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?
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?