Answer:
a) The velocity at which the water leaves the gun = 37.66 m/s
b) The volume rate of flow = (1.183 × 10⁻⁴) m³/s
c) The water hits the ground 18.64 m from the point where the water gun was shot.
Step-by-step explanation:
a) Using Bernoulli's equation, an equation that is based on the conservation of energy.
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
The two levels we are considering is just inside the water reservoir and just outside it.
ρgh is an extension of potential energy and since the two levels are at the same height,
ρgh₁ = ρgh₂
Bernoulli's equation becomes
P₁ + ½ρv₁² = P₂ + ½ρv₂²
P₁ = Pressure inside the water reservoir = 8 atm = 8 × 101325 = 810,600 Pa
ρ = density of water = 1000 kg/m³
v₁ = velocity iof f water in the reservoir = 0 m/s
P₂ = Pressure outside the water reservoir = atmospheric pressure = 1 atm = 1 × 101325 = 101,325 Pa
v₂ = velocity outside the reservoir = ?
810,600 + 0 = 101,325 + 0.5×1000×v₂²
500v₂² = 810,600 - 101,325 = 709,275
v₂² = (709,275/500) = 1,418.55
v₂ = √(1418.55) = 37.66 m/s
b) Volumetric flowrate is given as
Q = Av
A = Cross sectional Area of the channel of flow = πr² = π×(0.001)² = 0.0000031416 m²
v = velocity = 37.66 m/s
Q = 0.0000031416 × 37.66 = 0.0001183123 m³/s = (1.183 × 10⁻⁴) m³/s
c) If the height of gun above the ground is 1.2 m. Where does the water hit the ground?
The range of trajectory motion is given as
R = vT
v = horizontal component of the velocity = 37.66 m/s
T = time of flight = ?
But time of flight is given as
T = √(2H/g) (Since the initial vertical component of the velocity = 0 m/s
H = 1.2 m
g = acceleration due to gravity = 9.8 m/s²
T = √(2×1.2/9.8) = 0.495 s
Range = vT = 37.66 × 0.495 = 18.64 m
Hope this Helps!!!
