A forceful hydrant

In one second a cylinder of water (of radius 0.2 m, length 30 m, density 1000 kg/ $m^3$ and velocity 30 m/s) is turned from travelling upwards to travelling horizontally. There is therefore a loss of momentum of

$\pi \times 0.2^2 \times 30 \times 1000 \times 30$

in the vertical direction, and an equal gain in the horizontal direction.

By Newton's second law $\textbf{F}=\frac{\Delta m\textbf{v}}{\Delta t}$ these changes in momentum equal the two components of the force on the column of water. From his third law the forces exerted by the hydrant on the water are equal and opposite to those exerted by the water on the hydrant. These components form the sides of a square and the parallelogram law for vector addition then gives the total force on the hydrant as

$\sqrt{2} \times \pi \times 0.2^2 \times 30 \times 1000 \times 30=\boxed{159944}$ newtons.

At the outlet the velocity changes 90° angle. So change of modulus=√2v. Now at 1 sec out coming volume of water is=πr^2*v. At unit time thus we can get the change of momentum which is the force required.