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x’’(t)=g-Fd/m=(9.81metres/sec^2)-(k/m)(x’(t))

where x''(t)=du/dt and x'(t)=u=dx/dt

I enter this at http://www.wolframalpha.com/input/?i=

and I get the solution x(t)= -(c1me^(-kt/m))/k+c2+9.81(m/k)t

But

x(t=0)=0

x’(t=0)=0

x(t=0)=c1+c2

x’(t=0)=c1-c2

=>c1=c2=0

=>x(t)=9.81(m/k)t

and this says that it reached terminal speed right from the start of the fall, and it is not the solution mentioned at H.Young p.126. Where's the mistake?

where x''(t)=du/dt and x'(t)=u=dx/dt

I enter this at http://www.wolframalpha.com/input/?i=

and I get the solution x(t)= -(c1me^(-kt/m))/k+c2+9.81(m/k)t

But

x(t=0)=0

x’(t=0)=0

x(t=0)=c1+c2

x’(t=0)=c1-c2

=>c1=c2=0

=>x(t)=9.81(m/k)t

and this says that it reached terminal speed right from the start of the fall, and it is not the solution mentioned at H.Young p.126. Where's the mistake?

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