What Is the Speed of Water Leaving the Nozzle?

AI Thread Summary
The discussion revolves around calculating the speed of water leaving a garden hose nozzle pointed vertically upward. The initial attempt involved using kinematic equations, but confusion arose regarding the correct distance to use and the interpretation of time. Clarification was provided that the water takes 2 seconds to fall -1.5 meters from its starting point, emphasizing the importance of sign conventions in calculations. The correct approach involves ensuring that upward motion is considered positive while applying the equations accurately. Ultimately, understanding the problem's parameters is crucial for determining the correct water speed.
_Pistolkisses
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Homework Statement


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 2.0 s. What is the water speed as it leaves the nozzle?

Homework Equations


I'm guessing, d = v2t-(0.5)at^2
and v1^2= (v2^2)-2ad.

The Attempt at a Solution


I subbed in v2=0, a = -9.8, and t= 2.0 in the first equation to get distance.

And then subtracted 1.5 from the distance and used the new distance in the second equation (v1^2= v2^2 -2ad) to find velocity and came up with 18.84m/s.

It's not right because I had no idea which distance to use or how to find the right distance because the time value of 2.0 seconds is confusing me (the wording, anyways).

Should distance be -1.5m instead? :S
 
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_Pistolkisses said:
Should distance be -1.5m instead? :S

That's right. The water which leaves the nozzle with a speed of vi upward takes 2 secs to fall -1.5 m from the starting point, if up is taken as positive. Be careful about the signs of the other quantities.
 
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