A body of mass M is falling from a very high building. The drag force is given by F=kv^2 where k is a constant and v its velocity. How soon will it reach terminal velocity and what will the distance traversed be equal to?

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Answer 1 · 2017-12-09T13:35:31

#"The terminal velocity is approached exponentially but never"# #"reached exactly."#

#t = (1/2)sqrt(M/(kg)) ln[(1+sqrt(1-exp(-2 k h/M)))/(1-sqrt(1-exp(-2kh/M)))]#
#v_{max} = sqrt(M g/k)#
#v = sqrt[ (M g/k) (1 - exp(-2 k h / M))]#

#So if v -> v_{max} , " " then " " h -> oo , " and so " t -> oo#

These formulas are obtained if one solves the differential equation

#M {dv}/dt = M g - k v^2#

Note that

#h = "height (measured downwards) fallen in meter"#