The $\gamma$ here is some coefficient/constant (which has the units $\mathrm{kg\,s^{-1}}$). This formula is saying that the resultant force $F = \mathrm{d}v/\mathrm{d}t$ is equal to the force on the mass from gravity minus some force $\gamma v$ which is dependent on velocity.

To me, you are right, this does not make too much sense from a Newtonian standpoint. It would make more sense if the resistive force was written as $\lambda v^{2}$ (and the units of $\lambda$ amended accordingly), this could then represent an air/fluid resistance where $\lambda$ is some type of 'drag coefficient'.

I hope this helps.

The $\gamma$ here is some coefficient/constant (which has the units $\mathrm{kg\,s^{-1}}$). This formula is saying that the resultant force $F = \mathrm{d}v/\mathrm{d}t$ is equal to the force on the mass from gravity minus some force $\gamma v$ which is dependent on velocity.

To me, you are right, this does not make too much sense from a Newtonian standpoint (IMO) as this is representing a drag linearly dependent on velocity which is never the case in my experience. It would make more sense if the resistive force was written as $\lambda v^{2}$ (and the units of $\lambda$ amended accordingly for some arbitrary example case), this could then represent an air/fluid resistance where $\lambda$ is some type of 'drag coefficient'.

I hope this helps.