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sunnyceej
sunnyceej is offline
#6
Feb28-12, 09:58 PM
P: 14
So this gives me
[itex]\int\frac{dv}{mg-kv^{2}}[/itex] = [itex]\int\frac{dt}{m}[/itex]
with u substitution:
[itex]\sqrt{\frac{k}{mg}v}[/itex]([itex]\frac{1}{2}[/itex][itex]\int\frac{du}{1+u}[/itex] + [itex]\frac{1}{2}[/itex][itex]\int\frac{du}{1-u}[/itex]) = [itex]\int\frac{dt}{m}[/itex]

becomes

[itex]\sqrt{\frac{k}{mg}v}(\frac{1}{2}[/itex]ln[itex]\left|1+u\right|[/itex] + [itex]\frac{1}{2}[/itex]ln[itex]\left|1-u\right|[/itex]) = [itex]\frac{t}{m}[/itex]+c

then
[itex]\sqrt{\frac{k}{mg}v}(ln\sqrt{1+u}[/itex]+ln[itex]\sqrt{1-u}[/itex])= [itex]\frac{t}{m}[/itex]+c

[itex]\sqrt{\frac{k}{mg}v}(ln\sqrt{1-u^{2}}[/itex])=[itex]\frac{t}{m}[/itex]+c

and finally sub back in for u to get:
[itex]\sqrt{\frac{k}{mg}v}*ln\sqrt{1-\frac{k}{mg}v^{2}}[/itex]=[itex]\frac{t}{m}[/itex]+c

Is this right?