Answer:
v₁ = 1,606 10⁴ m / s
Explanation:
For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² +ρ g y₂
The pressure when the water comes out is the atmospheric pressure
P₁ = P_atm
The difference in height between the street and the nozzle on the stairs is
y₂-y₁ = 15 m
Now let's use the continuity equation
v₁ A₁ = v₂ A₁
The area of a circle is
A = π r² = π (d/2)²
v₁ π d₁²/ 4 = v₂ π d₂²/ 4
v₂ = v₁ d₁² / d₂²
Let's replace
P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²)²- v₁² ] + ρ g (y2-y1) = 0
P₂- P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1- (d₁/d₂)⁴]
v₁² [1- (d₁/d₂)⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2
Let's reduce the magnitudes to the SI system
d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m
d₁ = 2.02 in = 5.13 10⁻² m
Let's calculate
v₁² [1- (5.13 / 8.56) 4] = 449.538 10³ 10³/2 -9.8 15/2
v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸
v₁ = √ 2.2477 10⁸ /0.8710
v₁ = 1,606 10⁴ m / s
