Hoses, helicopters and rockets: force from a flow of mass

Firefighters brace themselves against their hoses, helicopters stay up by hurling air downwards, and rockets accelerate through empty space. None of these is "one object with one acceleration", so $F = ma$ is awkward. The momentum version of the second law handles all of them.

In a time $\Delta t$, a mass $\Delta m$ of water (or air, or exhaust gas) has its velocity changed by $\Delta v$. Its change in momentum is $\Delta m\,\Delta v$, so

$F = \frac{\Delta p}{\Delta t} = \frac{\Delta m}{\Delta t}\,\Delta v$

$\dfrac{\Delta m}{\Delta t}$ is the mass flow rate, in $\text{kg s}^{-1}$. Unit check: $\text{kg s}^{-1} \times \text{m s}^{-1} = \text{kg m s}^{-2} = \text{N}$. ✓

Sometimes you're given the jet's speed $v$, its cross-sectional area $A$ and the fluid's density $\rho$ instead. In a time $\Delta t$, a column of fluid of length $v\Delta t$ leaves the nozzle, with volume $Av\Delta t$, so

$\frac{\Delta m}{\Delta t} = \rho A v$

If the jet is brought completely to rest, $\Delta v = v$ and the force is $\rho A v \times v = \rho A v^2$.