(a) what is the acceleration of two falling sky divers (mass 132 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight? (b) after popping open the parachute, the divers descend leisurely to the ground at constant speed. what now is the force of air resistance on the sky divers and their parachute?

As per the question the mass of two falling sky drivers is 132 kg.

First we have to calculate their acceleration.

Whenever a body falls freely under gravity,its acceleration is acceleration due to gravity i.e g whose value is 9.8 m/s^2.

The earth pulls the object with a force equal to the weight of the body.

Hence the force gravity  F=W= mg   [ here m is mass of the body]

Here m =132 kg.

Hence force of gravity F= mg

                                        =132 kg ×9.8 m/s^2

                                        =1293.6 kg m/s^2

                                         =1293.6 N      [ here N[newton] is the unit of force.]

As per the question the air resistance is one fourth of weight of the bodies.

Hence air resistance F' =1/4 mg

Here F acts in vertically downward direction while F' acts in vertically upward direction.

Hence the net force acting on the particle is F-F'.

From Newton's second law of motion we know that net force is the product of mass and acceleration i.e  

                                     [Here a is the acceleration]

In the second question it has been told that they descend with uniform speed.hence acceleration of the two bodies will be zero.

 we know that F= ma

                           =m×0

                            =0 N

Hence they will not get any force when they will descend with a uniform speed.

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.