Explain why the acceleration becomes positive


A parachutist of mass 70 kg jumps from a plane at an altitude of 32 km above the surface of the Earth. Unfortunately, the parachute fails to open. (In the following parts, neglect horizontal motion and assume that the initial velocity is zero.)
(a) Calculate the time of fall (accurate to 1 s) until ground impact, given no air resistance and a constant value of g.
(b) Calculate the time of fall (accurate to 1 s) until ground impact, given constant g and a force of air resistance given by

F(v) = -c(subscript 2)*v*(magnitude v)

where c(subscript 2) is 0.5 in SI units for a falling man and is constant.
(c) Calculate the time of fall (accurate to 1 s) until ground impact, given c(subscript 2) scales with atmospheric density as
  c(subscript 2) = 0.5e^(-y/H)

where H = 8 km is the scale height of the atmosphere and y is the height above the ground. Furthermore, assume that g is no longer constant but is given by

g = 9.8/(1+y/R(subscript e) )^2 ms^-2

where R(subscript e) is the radius of the Earth and is 6370 km.
(d) For case (c), plot the acceleration, velocity, and altitude of the parachutist as a function of time. Explain why the acceleration becomes positive as the parachutist falls.

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Mechanical Engineering: Explain why the acceleration becomes positive
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