Let's call the gauge pressure P, the muzzle speed v, and the time of flight t.
First, we can find the time of flight using the vertical motion of the water. When the water hits the ground, its height is 0. Therefore:
0 = 0.85m + 1/2 * (-g) * t^2, where g is the acceleration due to gravity (9.8 m/s²)
t^2 = 2 * 0.85m / g
t^2 = 2 * 0.85m / 9.8 m/s²
t ≈ 0.416 s
Now, we need to find the muzzle speed. We know:
horizontal distance = 6.9m = muzzle speed * time of flight
v = 6.9m / 0.416 s
v ≈ 16.59 m/s
The Bernoulli's equation states that for a steady flow of an incompressible fluid:
P1 + 1/2 * ρ * v1^2 + ρ * g * h1 = P2 + 1/2 * ρ * v2^2 + ρ * g * h2
We have two points: one inside the reservoir (P1, v1=0, h1=0) and one at the muzzle (P2, v2=muzzle speed, h2=0.85m). So our equation becomes:
P + 0 + 0 = P2 + 1/2 * ρ * (16.59 m/s)^2 + ρ * g * 0.85m
We want to find P2 (the gauge pressure at the muzzle). The density of water (ρ) is 1000 kg/m³. Plugging in the values, we get:
P = P2 + 1/2 * 1000 kg/m³ * (16.59 m/s)^2 + 1000 kg/m³ * 9.8 m/s² * 0.85m
Now we can solve for P2:
P2 = P - (1/2 * 1000 kg/m³ * (16.59 m/s)^2 + 1000 kg/m³ * 9.8 m/s² * 0.85m)
P2 ≈ P - (137274.85 kg*m²/s² + 8303 kg*m²/s²)
P2 ≈ P - (145577.85 kg*m²/s²)
To find the gauge pressure, we need to convert the unit back to Pascal:
P2 ≈ P - 145578 Pa
So the gauge pressure at the instant the gun is fired is approximately 145578 Pascal.
A hand-pumped water gun is held level at a height of 0.85 m above the ground and fired. The water stream from the gun hits the ground a horizontal distance of 6.9 m from the muzzle. Find the gauge pressure of the water gun's reservoir at the instant when the gun is fired. Assume that the speed of the water in the reservoir is zero, and that the water flow is steady. Ignore both air resistance and the height difference between the reservoir and the muzzle.
I am completely lost at this problem.
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