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    Mar 7, 2007, 08:36 AM
    particle projected vertically
    a particle of mass m is projected vertically upwards from the origin, i.e. z(0)=0, with speed aU0 in a medium that produces a resistance to motion of mgv/U0 when the particles speed is v, where g is the acceleration due to gravity. Show that the particle will rise to a maximum height h in time T, where h=U0^2/g(a-log(1+a)), T=Uo/g(log(1+a)).

    If the particle returns to the point of projection with speed bU0(b<1) in a further time T', show that T'=-U0/g(log(1-b)).

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