Kinematics Water shooting out of hose

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SUMMARY

The discussion focuses on calculating the initial speed of water exiting a garden hose nozzle pointed vertically upward. The problem states that the nozzle is 1.5 meters above the ground and that the water strikes the ground 2.0 seconds after being turned off. The user derived an equation combining the kinematic equations for projectile motion, ultimately calculating the initial velocity (v_o) to be 5.275 m/s. Participants in the discussion emphasized the importance of clearly presenting equations and values to enhance understanding.

PREREQUISITES
  • Understanding of kinematic equations for constant acceleration
  • Basic algebra for solving equations
  • Knowledge of projectile motion principles
  • Familiarity with units of measurement in physics (meters, seconds)
NEXT STEPS
  • Study the derivation of kinematic equations in physics
  • Learn how to analyze projectile motion problems
  • Explore the concept of initial velocity in free-fall scenarios
  • Practice solving similar problems using different initial conditions
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Students studying physics, educators teaching kinematics, and anyone interested in understanding projectile motion and its applications in real-world scenarios.

hanlon
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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 1.5 above the ground. When you quickly turn off the nozzle, you hear the water striking the ground next to you for another 2.0 . find the water speed as it leaves the hose nozzel

GIANCOLI.ch02.p62.jpg

Homework Equations



v = v_o + at
x = x_o + v_o*t + (1/2)at^2
v^2 = v_o^2 + 2a(x - x_o)
v_a = (v + v_o)/2

where v_o is initial velocity x_o is initial distance v_a is average velocity

The Attempt at a Solution



dunno if I can actually do this but I derived my own equation adding the top half (the parabola area) and the bottom the part equal 1.5m

top

0 = 2v_o + 19.6t ( v_o + at) for both 2 halfs of the parabola

bottom

-1.5 = v_o*t + 4.9t^2

add top and bottom

0 = v_o*t - 4.9t^2 +1.5 + 2v_o + 19.6t

simplify

-v_o = (-4.9t^2 + 19.6t +1.5)/ (t+2)

solve for v_o with t = 2

v_o = 5.275 m/s

I don't know if I was mathematically correct to add those equations together, can anyone help please

Thank You.
 
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Sorry hanlon but your work is hard to follow. Use the subscript and superscript buttons provided above the input text field when posting, and any other necessary formatting to make it easier to read. You've also written down equations with values already plugged in without specifying which of the relevant equations you've used. It's hard to see where those values came from.

top

0 = 2v_o + 19.6t ( v_o + at) for both 2 halfs of the parabola

bottom

-1.5 = v_o*t + 4.9t^2

Where does the value 19.6 come from? Also, can you tell me how you got those equations you're using? Are they derived from the relevant, constant acceleration equations?
 
Last edited:
I, too, did not follow your derivation. Let's look at those equations again:
hanlon said:

Homework Equations



v = v_o + at
x = x_o + v_o*t + (1/2)at^2
v^2 = v_o^2 + 2a(x - x_o)
v_a = (v + v_o)/2

where v_o is initial velocity x_o is initial distance v_a is average velocity
Let's make a list of all the quantities that appear here:
t
x
x0
v
v0
a
(Since va is simply (x-x0)/t, I didn't bother listing it.)
Question for you: which of the quantities are given in the problem statement? Which quantity is being asked for? That information should help with choosing which equation (in your list above) will work out for this problem.
 

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