Answer:
a) v₂ = 32.83 m / s, b) y = 54.99 m
Explanation:
a) For this problem we must use Bernoulli's equation, we define the point as the surface of the water in the tank and point 2 is the point of the nozzle
P₁ + ½ ρ v₁² +ρ g y₁ = P₂ + ½ρ v₂² + ρ g y₂
As the nozzle is open to the outside it has the atmospheric pressure (P₂ = [tex]P_{atm}[/tex]) and the height y₂ = -4.00 m, since the tank is very large the water velocity inside the tank is very small, we will assume zero (v₁ = 0), replace
P₁ + ρ g (y₁-y₂) =[tex]P_{atm}[/tex] + ½ ρ v₂²
v₂² = 2 /ρ [(P₁ -[tex]P_{atm}[/tex]) + ρ g (0 - y₂)]
Let's calculate
v₂² = 2/1000 [(6.01 -1.013) 10⁵ + 1000 9.8 4]
v₂ = √ (2 10⁻³ [4,997 10⁵ + 3.92 10⁴]) = √ 1077.8
v₂ = 32.83 m / s
b) the height at which the water reaches can be found with kinematics at this point the speed is zero
Vy² = v₂² - 2 g y
vy = 0
y = v₂² / 2g
Let's calculate
y = 32.83²/2 9.8
y = 54.99 m
