Terminal Velocity - Falling with a Linear Drag Force

The total force on a particle that falls vertically under the influence of gravity and experiences a linear drag force is

$$F_z=-mg-bv_z,$$

where $m$ is the mass, $g=9.81$ m/s², and $b$ is the drag coefficient.

$z$

$t$

 $$z=z_0-\frac{mg}{b}t-\frac{m}{b}\left(v_{z0}+\frac{mg}{b}\right)\exp\left(-\frac{b}{m}t\right)+\frac{m}{b}\left(v_{z0}+\frac{mg}{b}\right)$$ $$v_z=-\frac{mg}{b}+\left(v_{z0}+\frac{mg}{b}\right)\exp\left(-\frac{b}{m}t\right)$$ $$a_z=-\frac{b}{m}\left(v_{z0}+\frac{mg}{b}\right)\exp\left(-\frac{b}{m}t\right)$$ $$F_z=-b\left(v_{z0}+\frac{mg}{b}\right)\exp\left(-\frac{b}{m}t\right)$$

$\frac{m}{b}=$ 1 [N]

$v_{z0}=$ 4 [N]

Consider a particle confined to move along the $z$-axis that falls vertically under the influence of gravity and experiences a linear drag force of the form, $F_{\text{drag}}=-bv_z$. The particle accelerates downward until the downward gravitational force is balanced by an upwards drag force. When this happens the total force is zero and the particle continues to travel downwards with a constant terminal velocity, $v_{\text{term}}=-mg/b$.

The total force on the particle is $F_z=-mg-bv_z$. The acceleration is,

$$a_z=-g-bv_z/m.$$

This can be written as a differential equation for the velocity,

$$\frac{dv_z}{dt}=-g-bv_z/m.$$

This differential equation can be solved numerically but it is also possible to determine the solution analytically. The solution to the differential equation is,

$$v_z=-\frac{mg}{b}+\left(v_{z0}+\frac{mg}{b}\right)\exp\left(-\frac{b}{m}t\right),$$

where $v_{z0}$ is the vertical velocity at time $t=0$. At $t=0$, the exponential factor is 1 and the velocity is $v_{z0}$. For times $t >> m/b$, the exponential factor approaches zero and the velocity becomes $-mg/b$.

The position can be determined by integrating the velocity once with repect to time,

$$z=z_0-\frac{mg}{b}t-\frac{m}{b}\left(v_{z0}+\frac{mg}{b}\right)\exp\left(-\frac{b}{m}t\right)+\frac{m}{b}\left(v_{z0}+\frac{mg}{b}\right).$$

The acceleration is the dervivative of the velocity,

$$a_z=-\frac{b}{m}\left(v_{z0}+\frac{mg}{b}\right)\exp\left(-\frac{b}{m}t\right).$$

The force is proportional to the acceleration,

$$F_z=-b\left(v_{z0}+\frac{mg}{b}\right)\exp\left(-\frac{b}{m}t\right)$$