Watering plants the hard way

For and ideal liquid, with no viscosity and friction to the wall of the hose, energy conservation holds. This yields the Bernoulli equation, $\frac {1}{2} \rho v^2 + \rho g h+p=const.$ , where $\rho$ is the density, $v$ is the velocity, $g$ is the acceleration of gravity, $h$ is the elevation and $p$ is the pressure. When the water is out of the hose the pressure is $p_0$ , the atmospheric pressure. When the water reaches the elevation of 3m its velocity is approximately zero. We can conclude that the constant in the equation is $const.=\rho g (3m)+p_0$ .

If Alice climbs up, the pressure in the hose will drop and the velocity of the out-coming water will also drop. These quantities can be determined from $\frac {1}{2} \rho v^2 + \rho g h+p=\rho g (3m)+p_0$ . For example, the pressure in the hose is $p_1= p_0+\rho g (3m) - \frac {1}{2} \rho v_1^2$ , where $v_1$ is the velocity of the water in the hose; the velocity of the water just after leaving the hose satisfies $v_2 = \sqrt{2 g (3m)}$ . Since the pressure cannot be less than $p_0$ , when she reaches $h=$ 3m, the only solution to the equation is $v=0$ and $p=p_0$ . The flow of water stops.

Water is not an ideal fluid and some of the energy is dissipated. The dissipation increases with the velocity of the flow. Some of the dissipation happens in the hose and the pipes bringing the water to the hose, some (most) happens in the nozzle, where the water speeds up before leaving the hose. Bernoulli’s equation is only approximately correct. For the water to reach 3m in the first job the pressure in the hose has to be a bit more than the value $p_1$ we calculated above. This is to compensate for the losses due to the viscosity. As Alice climbs up the ladder and the water flow gets slower, the losses will be smaller. There will be some flow (with low velocity and small loss of energy) even at 3m elevation.