A while ago, pinkybear posted the following: 1. The problem statement, all variables and given/known data A fire hose for use in urban areas must be able to shoot a stream of water to a maximum height of 35 m. The water leaves the hose at ground level in a circular stream 3.5 cm in diameter. What minimum power is required to create such a stream of water? Every cubic meter of water has a mass of 1.00*10^3 kg. 2. Relevant equations P=W/t W=Fd V(of cylinder)=pi*r^2*h 3. The attempt at a solution r=.0175m v=pi*(.0175)^2*35 =.033674 m^3 then, mass= 33.67kg w=(33.67)*g*35=11550 N but then I'm stuck because I don't know what t is... I tried v=v(0)+at, from which I got P=m*a^2*D/v(0) but now, I don't know how to get v(0)... I was not satisfied with the sole response given, since it involved unnecessary approximation (basically, run through the calculations for a small, but finite amount of water). Here is my calculus-based approach: It is useful to remember the following: Velocity: v=dy/dt Kinetic energy: K=1/2mv2 Power: P=dK/dt=1/2*d/dt(mv2) To use these, we need to calculate the velocity of the water on leaving the hose, which you can get by solving the following (remember that a is negative here): vf2 - v02 = 2a*Δy We can calculate the flow rate, which is the change in volume (capital V) over time, or dV/dt: dV/dt = dy/dt * A Then, using the density of water, we can convert this to the change in mass over time, dm/dt. We noted above that power is: P=1/2*d/dt(mv2). Since v is constant: P = 1/2v2 * dm/dt This is enough to answer the question. But we could go further and calculate the force of the water pushing back on the hose. We'll just use a couple more equations: Force (Newton's original definition): F=dp/dt Momentum: p=mv Again, v is constant. We substitute: F=d/dt (mv) = v * dm/dt Now we have the force, too! Pressure, then, is just dividing the force by the sectional area of the nozzle.