Difference between revisions of "Falling Body with Air Resist"

From apimba
Jump to navigation Jump to search
Line 40: Line 40:
  
 
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
 +
 +
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/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