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?

Question

1203 views around the world

You can reuse this answer
Creative Commons License

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"#