Answer:
The speed of the water as it leaves the nozzle is [tex]u =9.7 \ m/s[/tex]
Explanation:
From the question we are told that
The height of the nozzle above the ground is [tex]h = 1.5 \ m[/tex]
The time taken for the flow to stop is t = 1.8 s
According the second law of motion
[tex]h = ut - \frac{1}{2} gt^2[/tex]
making the initial velocity the subject
[tex]u = \frac{h + 0.5gt^2}{t}[/tex]
here [tex]g=9.8 \ m/s^2[/tex]
substituting value
[tex]u = \frac{1.5 + 0.5 * 9.8 * (1.8)^2}{1.8}[/tex]
[tex]u =9.7 \ m/s[/tex]
