A jet of water squirts out horizontally from a hole near the bottom of the tank shown in the figure below. If?

A jet of water squirts out horizontally from a hole near the bottom of the tank shown in the figure below. If the hole has a diameter of 3.20 mm, what is the height h of the water level in the tank?

Answer 1

First, find the time it takes the water to reach the ground from the instant it exits the hole, this can be found from kinematics :

Δy = v₀t + 0.5gt²

Since initial velocity (v₀) is zero vertically because the water comes out horizontally, this becomes :

Δy = 0.5gt²

Solved for t :

t = √[2Δy / g]
= √[(2 x -1.00m) /-9.80m/s²]
= 0.452s

Having the time allows us to calculate the speed of the water as it exits the hole, horizontally. Since velocity is distance over time :

v = Δx / t
= 0.600m / 0.452s
= 1.33m/s

With the speed, we may apply Bernoulli’s equation :

P₁ + 0.5ρv₁² + ρgy₁ = P₂ + 0.5ρv₂² + ρgy₂

The subscript 1 refers to the level of the water at the top of the tank, and 2 refers to the point of the hole in the tank. Assuming the tank is open to the atmosphere and the hole in the side is tiny compared to the open top, then P₁= P₂ = P₀ (atmospheric pressure). Also if the tank is large enough, the speed of the dropping water level in the tank is very small compred to the speed of the water leaving the tank, therefore, v₁= 0. Then Bernoulli’s equation reduces to :

ρgy₁ = 0.5ρv₂² + ρgy₂

The height h is the difference in the tank’s water level and the height above the ground of the small hole. That is :

h = y₁ – y₂

The simplified Bernoulli equation may be written as :

= y₁ – y₂ = h = v₂² / 2g

Then the height is :

h = (1.33m/s)² /(2 x 9.80m/s²)
= 0.0902m (or 9.025cm)

The diameter of the small hole is a red herring. Hope this helps.

Answer 2

Water Squirts

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