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Unformatted text preview: à infty, v T = mg/k (1) v T =/= \infty (2) v T is independent of V0 (the initial speed) (3) Even if v0 =0, we can reach the positive speed (4) if k=0 (frictionless), v T à \infty (5) if k=\infty, v T à Resistance Proportional to Square of Velocity (high speed situation…) Equation can be written as m v’ = kvv – mg whose solution is For Upward move: ( v = v because v is positive) V’ = g – r v^2 We found v(t) = sqrt(g/r) tan {C1 – t sqrt(rg) } where C1 = arc tan {v0 sqrt(r/g) } For Downward move: ( v = v because v is negative) V’ = g + r v^2 We found v(t) = sqrt(g/r) tan h {C2 – t sqrt(rg) } where C2 = arc tanh {v0 sqrt(r/g) } Where tanh (x) = sinh (x)/cosh (x) = ½ {e^x – e^(x)}/ ½ {e^x + e^(x)} Remarks: (1) If v0=0, we can easily see C2=0 so v(t) = sqrt(g/r) tan h {t sqrt(rg) } (2) Terminal speed is t=infty, we have v(t=infty) = sqrt(q/r). This is the terminal speed....
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 Summer '08
 Staff
 Force, Mass, BVP, terminal speed, Resistance Proportional

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