Difference between revisions of "Falling Body with Air Resist"
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when t goes to infinity, e<sup>-kt/m</sup> goes to 0 so the terminal velocity = mg/k | when t goes to infinity, e<sup>-kt/m</sup> goes to 0 so the terminal velocity = mg/k | ||
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+ | The value of k depends on mass and shape, so it can be determined by measuring terminal velocity: | ||
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+ | average terminal velocity of an 80kg person = 148mph = 66m/s so k = 80kg(9.8m/s<sup>2</sup>)/66 m/s = 11.8 kg/sec |
Revision as of 07:38, 5 May 2020
Ftotal = Fgrav - Fair
ma = mg - kv where k units are kg/sec
mdv/dt = mg - kv
put into standard form:
mdv/dt + kv = mg
dv/dt + vk/m = g <----Standard form
find u = e∫k/mdt = ekt/m
multiply by u:
ekt/mdv/dt + ekt/mvk/m = gekt/m
use product rule:
d/dt(vekt/m) = gekt/m
integrate both sides:
vekt/m + C1 = mg/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
The value of k depends on mass and shape, so it can be determined by measuring terminal velocity:
average terminal velocity of an 80kg person = 148mph = 66m/s so k = 80kg(9.8m/s2)/66 m/s = 11.8 kg/sec