# More constant acceleration questions

1. Jan 12, 2006

### kylera

Because there's no solution, and my book rated this problem as one of the tougher ones compared to the others, I have no idea whether I did this right, but my gut feeling is that it's wrong because I solved it waaaay too quickly.

Question: Suppose you adjust your garden hose nozzle for a hard stream of water. You point the nozzle vertically upoward 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?

My solution: Use the formula that determines position with time, initial velocity and acceleration:
x = x(initial) + v(initial)*time + .5*a*t^2
t = 2.0s, a = -9.8m/s^2
-1.5 = 0 + 2v - 2*9.8
v(initial) = 9.1 m/s

Any pointers?

2. Jan 12, 2006

### lightgrav

okay, I got 9.05 m/s .
maybe the author thought it was "difficult" because he did not include one exactly like it (solved as an example).
Authors don't distinguish between "challenging", as opposed to "long" or "tedious".