Answer:
[tex]v_o = 30.5 m/s[/tex]
Explanation:
Height of the hose pipe is given as
[tex]y = 4.5 m[/tex]
horizontal distance is given as
[tex]x = 25 m[/tex]
angle of elevation is given as 35 degree with the horizontal
now we know that the horizontal distance moved is given as
[tex]x = (v_ocos35) t[/tex]
[tex]y = 4.5 + (v_o sin35) t - \frac{1}{2}(9.8)t^2[/tex]
now we have
[tex]25 = v_o(0.82)t[/tex]
also we have
[tex]0 = 4.5 + 0.57 v_o t - 4.9 t^2[/tex]
[tex]0 = 4.5 + (0.57)(0.82) - 4.9 t^2[/tex]
[tex]t = 1.00 s[/tex]
so we have
[tex]v_o = 30.5 m/s[/tex]