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?
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