91Ó°ÊÓ

It is given that,

$$\begin{gathered} massofperson\left(M\right)=80kg \\ massofparachute\left(M\right)=5kg \\ a=2.5m/s^2 \end{gathered}$$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Thus, using the free body diagramfor parachute and system of parachute and person, the force can be calculated.

Formula:

According to the Newton’s second law of motion,

$F_{net}=\underset{}{∑Ma}$

where, is the net force, Mis mass and a is an acceleration.

From first FBD:

W - F = ( M +m ) x a

(M + m )g - F = (M +m) x a

(85 x 9.8) - F = 85 x 2.5

By solving for F

F = 620 N

Hence, The upward force on the open parachute from the air is 620 N

From second FBD:

$$\begin{gathered} W-F_p=Ma \\ (80x9.8)-F_p=80x2.5 \\ Hence, \\ {F}_p=584N. \\ Hence,Thedownwardforceontheparachutefromthepersonis584N \end{gathered}$$