# Kinematic question from giancoli

1. Aug 6, 2011

### nobelium102

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 (the hose is 1.5 m off the ground). When you quickly move the nozzle from the vertical 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

?

3. The attempt at a solution

my attempt s = ut+1/2 at^2
-1.5 = u(2) + 1/2(-9.8)(4)
u = 9.05 m/s

and it is right
but the question is that why is the acceleration still -9.8m/s^2 when the water is going up and then falling down???
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Aug 6, 2011

### Delphi51

The acceleration is always downward so it is negative. It means the upward speed decreases steadily at a rate of 9.8 m/s every second. Initially the upward speed is 9.05 m/s. After 1 second, it is 9.05-9.81 = -.76 m/s. That is just after the maximum height where the speed is zero. After two seconds, the upward speed is -.76 - 9.81 = -10.57 m/s, which is the speed just before hitting the ground.

3. Aug 6, 2011

### nobelium102

oh man i didn't understand at first but i think i kind of get it

you are saying that acceleration has decreased so much that it became negative direction?
right?
hope this is right

4. Aug 6, 2011