Help with a kinematics question

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The discussion revolves around solving a kinematics problem involving a garden hose nozzle pointed vertically upward. The initial height of the nozzle is 1.5 meters, and the water takes 2 seconds to hit the ground after being released. The equation used is x(t) = height + Vo * t - 1/2 * a * t^2, where the user initially struggles with two variables. After clarification, it is determined that the position at t=2 seconds is 0, leading to the calculation of the initial velocity as 9.25 m/s. The conversation concludes with the user successfully finding the solution with assistance.
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Homework Statement


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?

Homework Equations


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

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) - 20Now 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.
 
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Hi danhamilton,

danhamilton said:

Homework Statement


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?


Homework Equations


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

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.

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.
 
I guess I meant initial position when I said height. So would that make x(t) 0?
 
Ok, so then
0 = 1.5 + v(2) - 20

18.5 = v(2)

So the initial velocity would be 9.25 m/s?
 
You got it.
 
Awesome, thank you guys so much for the help!
 

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