The upward force on the bungee jumper is expressed using the following linear relationship:
Fu = -c'v
where c' = a first-order drag coefficient (kg/s) and the downward force on the bungee jumper
is expressed using the following relationship:
FD = mg
where g is the acceleration due to gravity (9.81 m/s2).
a) Use calculus to identify which of the following best describes the closed-form solution
for the case where the bungee jumper is initially at rest (v = 0 at t = 0).

The problem is an application of the Ordinary Differential Equations in physics. We are given the functions for the two forces acting on an object. We apply the second Newton's law to find the dynamics conditions as follows

[tex]\displaystyle F_u+F_D=ma[/tex]

Knowing that the acceleration is the derivative of the velocity, thus

[tex]\displaystyle a=\frac{dv}{dt}=v'[/tex]

We know

[tex]\displaystyle F_u=-c'v[/tex]

and also

[tex]\displaystyle F_D=m\ g[/tex]

Replacing in the first equation:

[tex]\displaystyle -c'\ v+mg=mv'[/tex]

Rearranging

[tex]\displaystyle mv'+c'\ v=mg[/tex]

[tex]\displaystyle v'+\frac{c'}{m}\ v=g\ \ \ \ \ .....[eq\ 1][/tex]

This is a linear non-homogeneous ordinary differential equation. We'll solve it by finding the homogeneous solution first which takes the form

[tex]v_h(t)=ke^{Dt}[/tex]

It takes us to solve the characteristic equation:

[tex]\displaystyle mD+c'=0[/tex]

Solving for D

[tex]\displaystyle D=-\frac{c'}{m}[/tex]

The homogeneous solution is

[tex]\displaystyle v_h=Ke^{-\frac{c}{m}t}[/tex]

The particular solution depends on the form of the independent term located in the right side of the equation. In this case, it's a constant, thus the particular solution has the form

[tex]\displaystyle v_p=A[/tex]

Where A is a constant we must find, by replacing it in the original differential equation. We compute

[tex]\displaystyle v_p'=0[/tex]

Replacing in (Eq 1):

[tex]\displaystyle 0+\frac{c'}{m}A=g[/tex]

We find the value of A

[tex]\displaystyle A=\frac{mg}{c'}[/tex]

The complete solution for v is

[tex]\displaystyle v=vh+v_p[/tex]

[tex]\displaystyle v=k_e^{-\frac{c'}{m}t}+\frac{mg}{c'}[/tex]

We have the initial condition

[tex]\displaystyle v(0)=0[/tex]

That gives us the value of k

[tex]\displaystyle 0=k+\frac{mg}{c'}[/tex]

[tex]\displaystyle k=-\frac{mg}{c'}[/tex]

Thus, the solution for v is

[tex]\displaystyle v=\frac{-mg}{c'}\ e^{-\frac{c'}{m}t}}+\frac{mg}{c'}[/tex]

[tex]\displaystyle v=\frac{mg}{c'} \left (1-e^{-\frac{c'}{m}t}\right )[/tex]

The option marked iv. is the correct answer