# Water being hosed into the air.

## Homework Statement

Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle
vertically upward at a height of 1.5m above the ground. When you quickly move the nozzle away
from the vertical, you hear the water striking the ground next to you for another 2.0s. What is
the water speed as it leaves the nozzle?

## Homework Equations

Δx =Vo t + 1/2 at^2

## The Attempt at a Solution

Δx =Vo t + 1/2 at^2
-1.5m = (2s) Vo + (1/2) (-9.8 m/s^2) (2s)^2
-1.5m = (2s) Vo - 19.6 m
Vo = 9.05 m/s

Now the above solution seems to be correct, but I am not sure why t should = 2s.
I ended plugging 2 sec. in without really thinking about it, but now I am not able to
spatially visualize what is going on with the water.

Any help appreciated.

## Answers and Replies

This is just a very standard linear kinematics question that's been verbally tarted up. The point is that once the hose is moved away from vertical the last 'drop' of water to leave the nozzle hits the ground 2s later. So you're right to use 2s as the time in this case. This problem is mathematically identical to the usual rock/ball/textbook/whatever being thrown into the air.

Hello!

Thank you, that makes sense now. :)