The terminal velocity of a 2.8 *10-5 kg raindrop is about 6 m/s. (a) Assuming a

Question

The terminal velocity of a 2.8 *10-5 kg raindrop is about 6 m/s. (a) Assuming a drag force FD = -bv, determine the value of the constant b.
b=4.57e-5 correct answer... but then the next part, which I already asked someone on cramster who got is wrong, is very difficult and may involve ln and e in calculous or something.
the question is,(b) Determine the time required for such a drop, starting from rest, to reach 63 percent of terminal velocity.
No one seems to know the answer. it is not =.410
PLEASE help me with this. It is very difficult.

Explanation / Answer

let t=0 when vx=0. we have that dv/dt=g-bv/m. (this is the equation of motion) integrate this we have. dv/(g-bv/m)=dt. so dv/(mg/b-v) = dt*b/m so we have ln(v-mg/b)=-t*b/m+C. let v=0 when t=0. so C=ln(-mg/b). ln(v-mg/b)=-t*b/m + ln(-mg/b) so ln((v-mg/b)/-(mg/b))=-t*b/m e^(-tb/m)=(v-mg/b)/(-mg/b)=(mg/b-v)/(mg/b) (mg/b)*(1-e^(-tb/m))=v where mg/b is the terminal velocity v0. so with v=0.63v0 we have that 1-e^(-tb/m)=0.63 so t=0.61(s)

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