If an object of mass m| is dropped from rest, one model for its speed v| after l

Question

If an object of mass m| is dropped from rest, one model for its speed v| after l| seconds, taking air resistance into account is: v = mg/c(1 -e^(-ct/m))| Where g| is the acceleration due to gravity and c| is a positive constant. (In chapter 9 we will be able to deduce this equation from the assumption that the air resistance is proportional to the speed of the object. (a) Calculate lim v|. Your answer may depend on c|, g| and m|. (b) For fixed t|, use L'Hospital's rule to calculate lim v|. Your answer may depend on c|, g|, and t|. gt

Explanation / Answer

given v =(mg/c)(1-e-ct/m)

a)limt->v

=limt->(mg/c)(1-e-ct/m)

=(mg/c)(1-e-c()/m)

=(mg/c)(1-e-)

=(mg/c)(1-0)

=(mg/c)

limt->v=mg/c

b)limm->v

=limm->(mg/c)(1-e-ct/m)

=limm->(g/c)(1-e-ct/m)/(1/m)

we get 0/0 form apply l hospitals rule, differentiate numerator and denominator with respect to m

=limm->(g/c)(0-(ct/m2)e-ct/m)/(-1/m2)

=limm->(g/c)((ct)e-ct/m)

=limm->(gt)e-ct/m

=(gt)e-ct/

=(gt)e0

=(gt)

limm->v =gt

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