Answer:
The initial velocity of the water from the tank is 5.42 m/s
Explanation:
By applying Bernoulli equation between point 1 and 2
[tex]\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L[/tex]
At the point 1
P₁=0 ( Gauge pressure)
V₁= 0 m/s
Z₁=3 m
At point 2
P₂=0 ( Gauge pressure)
Z₂= 0 m/s
[tex]h_L=1.5\ m[/tex]
Now by putting the values
[tex]\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L[/tex]
[tex]Z_1-h_L=\dfrac{V_2^2}{2g}[/tex]
[tex]3-1.5=\dfrac{V_2^2}{2\times 9.81}[/tex]
[tex]V_2=\sqrt{2\times 1.5\times 9.81}\ m/s[/tex]
V₂= 5.42 m/s
The initial velocity of the water from the tank is 5.42 m/s