The differential equation `dv text[/]dt=g` is the inverse of a problem we have studied already: If a function is linear, then its average rate of change is constant (the slope of the linear graph), so its instantaneous rate of change is also constant. Having studied the inverse problem, we know solutions to the differential equation: Any linear function with slope `g`, say `v=gt+C`, satisfies the differential equation. Substituting `v=0` and `t=0` simultaneously, we see that the initial condition is satisfied if `C=0`. Thus, `v=gt` is one solution of the initial value problem
`(dv)/(dt)=g` with `vtext[(]0text[)]=0`.
We assumed that each initial value problem has only one solution, so the solution must be `v=gt`. If our units are metric, this means that the velocity at time `t` is `9.807 t` meters per second. If the units are English, then `vtext[(]t text[)]` is `32.17 t` feet per second. Notice that, by choosing acceleration to be positive in the downward direction, we have also made "down" the positive direction for velocity as well.
In general, we have `vtext[(]0text[)]=g*0+C`, so `C` is the initial velocity. If `vtext[(]0text[)]=10text[,]` then `v=gt+10` at all times `t` until impact. The physical interpretation in this case is that the object was thrown downward with an initial speed of `10` units. Of course, it matters whether those units are feet per second or meters per second, but the formula works either way, as long as the proper value is given to `g`. Similarly, an initial velocity of `-10` means the object is thrown upward with a speed of `10` units, and its velocity thereafter will be `v=gt-10`.