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
![=\frac{1}{4} *1293.6N](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B4%7D%20%2A1293.6N)
![=323.4 N](https://tex.z-dn.net/?f=%3D323.4%20N)
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'.
![F_{net} =1293.6N -323.4N](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D1293.6N%20-323.4N)
![=970.2 N](https://tex.z-dn.net/?f=%3D970.2%20N)
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]
![a =\frac{F_{net} }{m}](https://tex.z-dn.net/?f=a%20%3D%5Cfrac%7BF_%7Bnet%7D%20%7D%7Bm%7D)
![= \frac{970.2}{132} m/s^2](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B970.2%7D%7B132%7D%20m%2Fs%5E2)
![=7.35 m/s^2](https://tex.z-dn.net/?f=%3D7.35%20m%2Fs%5E2)
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.