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2 years ago

A jet of water squirts out horizontally from a

Physics

1 answer:

Triss [41]2 years ago

6 0

Answer:

9.54 cm

Explanation:

The water is in free fall after it leaves the tank. The time it takes to land is:

Δy = v₀ t + ½ at²

1.23 m = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.501 s

Therefore, the horizontal velocity is:

v = 0.685 m / 0.501 s

v = 1.37 m/s

Using Bernoulli equation, where point 1 is the surface of the water and point 2 is the exit of the tank:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since P₁ = P₂, and v₁ = 0:

ρgh₁ = ½ ρ v₂² + ρgh₂

gh₁ = ½ v₂² + gh₂

gh = ½ v₂²

h = v₂²/(2g)

Plugging in:

h = (1.37 m/s)² / (2 × 9.8 m/s²)

h = 0.0954 m

h = 9.54 cm

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