A freely falling spherical rain drop gathers moisture while (maintaining its spherical shape all the way) from the atmosphere at a rate $$\frac{dm}{dt}=kt^2$$, where $$t$$ is the time and $$m$$ is the instantaneous mass of the drop. The constant is $$k = 12\ \text{g}/\text{s}^3$$. If the drop, of initial mass $$m_0 = 2\ \text{g}$$, starts falling from rest, the instantaneous velocity exactly after $$5\ \text{s}$$ shall be, ignoring air friction and air buoyancy.