Question: Two questions on conversion

To check one of the MythBusters TV episodes, i.e., the fall of a mannequin (80 Kg) from a plane at an altitude of 1200 m in 31 s, with Maple13 (Windows Vista) I solved the following differential equation with initial conditions:

> de:= m*(D@@2)(x)(t) = m*g - k*(D(x)(t))^2:

> ini:= (x(0) = 0, D(x)(0) = 0):


> X:= unapply(rhs(expand(dsolve({de,ini}))),t);

        X := proc (t) options operator, arrow; -m^(1/2)*g^(1/2)*t/k^(1/2)-m*ln(2)/k+m*ln(exp(2*g^(1/2)*k^(1/2)*t/m^(1/2))+1)/k end proc

1) Posing: V0 = sqrt(m * g / k),  T = sqrt(m/(g*k)) one has  V0*T = m/k. I want to have:

   > Xxb:= t -> V0*(- t + T * ln((exp(2*t/T) + 1)/2) ); # m.

 How to obtain this equivalent equation  Xxb(t), without retyping the equation of  X(t) ? With subs , convert or simplify  applied to  X(t), Maple 13 gives error messages.

2) Taking the derivative of  Xxb(t), one find :

     V := proc (t) options operator, arrow; V0*(exp(2*t/T)-1)/(exp(2*t/T)+1) end proc.

How to transform it to V = V0 tanh(t/T) ?

Again, subs , convert or simplify  seem not working, even with  assume(t, real), assume(T, real) ! I know simplify is a difficult task, but Maple should recognize a tanh !

Thank you in advance for any suggestion.

( Note that with m = 80 Kg, g = 9.81 m/s^2 and k = 0.267482 Kg/m, which correspond to a speed limit of V0 =195 km/h, on find t(1200m) = 26 s, instead of 31 s ).