Difference between revisions of "Falling Body with Air Resist"

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v(t) = (mg/k)(1-e<sup>-kt/m</sup>)
 
v(t) = (mg/k)(1-e<sup>-kt/m</sup>)
 +
 +
when t goes to infinity, e<sup>-kt/m</sup> goes to 0 so the terminal velocity = mg/k

Revision as of 07:06, 5 May 2020

Ftotal = Fgrav - Fair

ma = mg - kv where k units are kg/sec

mdv/dt = mg - kv

put into standard order:

mdv/dt + kv = mg

dv/dt + vk/m = g

find u = e∫k/mdt = ekt/m

multiply by u:

ekt/mdv/dt + ekt/mvk/m = gekt/m

d/dt(vekt/m) = gekt/m

integrate both sides:

vekt/m + C1 = gm/kekt/m + C2

solve for v:

v = mg/k - (C1 + C2)/ekt/m

v = mg/k - Ce-kt/m

If at time 0, the body is at rest then v(0) = 0 = mg/k - C so C = mg/k

therefore v = mg/k - mg/ke-kt/m

or

v(t) = (mg/k)(1-e-kt/m)

when t goes to infinity, e-kt/m goes to 0 so the terminal velocity = mg/k