# Why do we use $m\frac{dv}{dy} = -b(v-v_{ter})$ while determining how the terminal velocity is changing for an object falls down in linear drag force?

Why do we use $$m\frac{dv}{dy} = -b(v-v_{ter} )$$ while determining how the terminal velocity is changing for an object falls down in linear drag force . I was Jr Taylor's classical mechanics. In the one dimensional or y directional projectile in linear drag proportional to (instantaneous velocity)^1 . I found this equation. Here drag force is given by $F_{drag} = -b(v - v_{ter})$, $v_{ter}$= terminal velocity. Now while in this equation why do we rule out gravitational force and why? Here $b$ is the constant which is the ratio of drag force and instantaneous velocity.

• What is b here? – Steeven Aug 23 '18 at 9:17
• @Steeven Here b is the constant which is the ratio of drag force and instantaneous velocity. – Nobody recognizeable Aug 23 '18 at 9:18

Consider the downward direction to be positive. The equation of motion is, $$m \frac {dv}{dt} = mg -bv$$, where $$bv$$ is the upward drag force.
The net force will be zero when $$v = v_{ter} =mg/b$$ Substituting back, we get $$F_{downward} =- b(v-v_{ter})$$.