# Help with a kinematics question

1. Sep 23, 2008

### danhamilton

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 1.5m 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 2.0 seconds. What is the water speed as it leaves the nozzle?

2. Relevant equations
I'm assuming you have to use the formula
x(t)=height + Vo * t - 1/2 * a * $$t^2$$

3. The attempt at a solution
So I plugged in the values given, and came up with this.

x(t)=1.5 + v(2) -1/2 * 10 * $$2^2$$

x(t)=1.5 + v(2) - 20

Now I'm kind of stuck though. It seems to me I have two variables, and nothing to do with them. Any help would be greatly appreciated.

2. Sep 24, 2008

### alphysicist

Hi danhamilton,

In your equation, you have labeled the first term on the right side as 'height'. That's true, but it's not just any height--it's the initial position (at t=0).

That's important because then on the left side x(t) is the position of the object at some time t. You've chosen t=2 (on the right side of the equation), so x(t) is the position at t=2. What would that be? Once you have that I believe you can solve for the initial speed.

3. Sep 24, 2008

### danhamilton

I guess I meant initial position when I said height. So would that make x(t) 0?

4. Sep 24, 2008

### Redbelly98

Staff Emeritus
Yes.

5. Sep 24, 2008

### danhamilton

Ok, so then
0 = 1.5 + v(2) - 20

18.5 = v(2)

So the initial velocity would be 9.25 m/s?

6. Sep 24, 2008

### Redbelly98

Staff Emeritus
You got it.

7. Sep 24, 2008

### danhamilton

Awesome, thank you guys so much for the help!