A jet of water from a 25mm diameter nozzle is directed vertically upwards...?

Question

A jet of water from a 25mm diameter nozzle is directed vertically upwards.Assuming that the jet remains circular and neglecting any loss of energy what is the diameter of the jet at a point 4.5m above the nozzle if the velocity with which the jet leaves the nozzle is 12 m/s ?

Answer 1

What a ridiculous question. After all, what is going to happen to the water when it reaches its peak elevation and starts falling through the column of water moving up?

But, what the question is supposed to be about is conservation of mass in a flow.

At the nozzle, you have a water velocity of Vn with a cross-sectional area An of pi r^2 for a flow rate of Vn An.

At a height h above that point, you have to have the same total flow rate, but the velocity is lower so the cross-sectional area must be greater.

Consider a mass of water m. At the nozzle it has no potential energy and a kinetic energy of (1/2) m Vn^2.

After it has risen a distance h, it has neither gained nor lost energy, but it has a potential energy of mgh, where g is the gravitational acceleration.

Therefore, its new velocity Vh must be such that:

(1/2) m Vn ^2 = m g h + (1/2) Vh ^2

This allows you to compute Vh so you can solve:

Vn An = Vh Ah for Ah.

From Ah you can find the diameter of the jet.