Answer:
y₂= 2.08 m
Explanation:
This problem should use the fluid continuity equation and kinematics to find the speeds.
Let's start by finding the initial velocity (vo) so that the water reaches the height (y1 = 0.13 m), with kinematics
vf² = v₀² - 2 g y₁
0 = v₀² - 2 g y₁
v₀ = √ 2g y₁
v₀ = √ (2 9.8 0.13)
v₀ = 1,596 m / s
Now let's use the fluid continuity equation
A₁ v₁ = A₂ v₂
The initial speed the speed with the largest diameter
v₁ = v₀ = 1,596 m / s
Let's look for the areas
r = d / 2
r₁ = 20/2 = 10 mm = 10 10⁻³m
r₂ = 10/2 = 5 mm = 5 10⁻³ m
A = pi R²
A₁ = π (10 10⁻³)²
A₁ = π 10⁻⁴ m²
A₂ = π5 10⁻³)²
A₂ = π 25 10⁻⁶ m²2
We cleared the exit speed (v2) by reducing the diameter (d2)
v₂ = v₁ A₁ / A₂
v₂ = 1,596 π 10⁻⁴ / π 25 10⁻⁶
v₂ = 6.38 m / s
Now we find the height for this speed again
vf² = v₂² - 2g y₂
0 = v₂² - 2 g y₂
y₂ = v₂² / 2g
y₂= 6.38² / (2 9.8)
y₂= 2.08 m