1. The problem statement, all variables and given/known data Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upward at a height of 1.5 m above the ground. When you quickly turn off the nozzle, you hear the water striking the ground next to you for another 2.0 s. What is the water speed as it leaves the nozzle? 2. Relevant equations vavg = Δx/ Δt vfinal2 =vinitial2 + 2ax 3. The attempt at a solution From the vavg equation: vg = 1.5/2t + 10t v0 = 1.5/2t - 10t From the other equation http://www4a.wolframalpha.com/Calculate/MSP/MSP10431h9ic8g7736e702600003bb73a2eg362c825?MSPStoreType=image/gif&s=34&w=238.&h=35. [Broken] http://www4a.wolframalpha.com/Calculate/MSP/MSP38881h9ic4fc19b732c400004f1gdedhga6cgg3b?MSPStoreType=image/gif&s=34&w=38.&h=18. [Broken] http://www4a.wolframalpha.com/Calculate/MSP/MSP38851h9ic4fc19b732c400002b4de3h0dg0716h1?MSPStoreType=image/gif&s=34&w=49.&h=18. [Broken] v0 was (1.5/2t - 10t). So it becomes 0.75 - 10 = -9.25. I think my answer is correct because I've come across the value of 9.25 on some other topics too. I'm not asking for a value correction. But what I found is minus 9.25 while others claim it is plus 9.25. Is this because of the direction of the water? Also, what about the spots I wrote (1-t). Have they simply become zero?