What Is the Speed of Water Leaving the Nozzle?

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SUMMARY

The discussion focuses on calculating the speed of water leaving a garden hose nozzle pointed vertically upward. The key equations used are d = v2t - (0.5)at² and v1² = v2² - 2ad. The correct calculation shows that the water speed as it leaves the nozzle is 18.84 m/s, derived from a distance of -1.5 m and a time of 2.0 seconds. Participants emphasize the importance of correctly interpreting the distance and the signs of the quantities involved in the equations.

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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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