# 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 $$F_{downward} =- b(v-v_{ter})$$.

Note that this is the net force (so it becomes zero when the body attains terminal velocity)., and not just the air drag (which is exactly equal and opposite to gravitational force at terminal velocity, making the net force zero)