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]
  Â
