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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 2.1s.
What is the water speed as it leaves the nozzle?

Respuesta :

Answer:

[tex]v_y = 9.57 m/s[/tex]

Explanation:

As we know that it will follow the kinematics

[tex]y - y_o= v_y t + \frac{1}{2}at^2[/tex]

now we will have

[tex]0 - 1.5 = v_y (2.1) - \frac{1}{2}(9.8) (2.1)^2[/tex]

here we can now solve above equation for speed vy

[tex]-1.5 = 2.1 v_y -21.61[/tex]

so we have

[tex]21.6 - 1.5 = 2.1 v_y[/tex]

[tex]v_y = \frac{20.11}{2.1}[/tex]

[tex]v_y = 9.57 m/s[/tex]

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