Suppose at the start of the parabola the aircraft has
 initial altitude H meters,
 speed Vo m/s,
 and is moving upward at an angle to the horizontal M degrees.

 If the desired gravitational force is G m/s^2
 then the aircraft must accelerate vertically at (9.81-G) m/s^2 .
 The horizontal component of the speed is constant at Vx(t) = Vo*cosM,
 so that the horizontal co-ordinate after t seconds is X(t) = Vo*cosM*t .

 Initially the vertical component of the speed is Vy(t=0) = Vo*sinM .
 The vertical component of the speed is Vy(t) =  Vo*sinM -(9.81-G)*t .
 At the peak vertical speed is zero, and occurs at t~ = Vo*sinM/(9.81-G)  .

 The vertical co-ordinate is Y(t) =  H + Vo*sinM*t - (9.81-G)*(t^2)/2 .
 The maximum height attained is
 Y(t~) = H+Vo*sinM*[Vo*sinM/(9.81-G)] - (9.81-G)*[Vo*sinM/(9.81-G)]^2/2 .
 By symetry it takes t~ seconds to go from the peak to the base altitude H.

 Greatest altitude and time at reduced gravity occur if the initial angle is 90
 degrees, but steep angles introduce other problems.  For a given initial
 airspeed and initial angle, a larger perceived gravity (as of Mars) has longer
 reduced gravity time than a smaller perceived gravity (Luna or freefall).

 There must be sufficient airspace below the base height to accelerate and
 maneuver into the parabola, and recover from the parabola.  At the peak there
 must be sufficient air density  and indicated air speed for the control surfaces
 to function.  The indicated air speed may be significantly less than the true
 air speed, and the pilot must compensate for winds that vary with altitude.

 I am not sure that the perceived gravity will be perpendicular to the aircraft
 floor.  For free fall that would not matter.                              -MBM

 Example
    H   = 6000 m            base height
    Vo  = 200 m/s           initial speed
    M   = 30 deg            initial flight angle
          3.71 m/s^2        desired perceived gravity
          6.10 m/s^2        needed vertical acceleration
   X(t) = 173 m/s           horizontal component of speed
    t~  = 16.4 s            time to max height
  Y(t~) = 6820 m            max height