Question

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 {\rm m} 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 1.4s What is the water speed as it leaves the nozzle?

Answer 1

The water speed it leaves the nozzle is

v₁ = 5.78 m / s

Step-by-step explanation:

The acceleration of the water after it leaves the nozzle is just due to gravity so:

a = g = -9.8 m/s² , s = 1.5 m , t = 1.4 s

Using the equation of the law of Newton that describe the motion as a:

s = v₁*t + ¹/₂*a*t²

Solve to v₁

v₁ = [ ¹/₂*a*t² - s ] / t

v₁ = [¹/₂ * 9.8 m/s² * 1.4²s - 1.5 m] / 1.4 s

v₁ = 5.78 m / s