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.
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.
No problem! Let's break down the problem step by step.
First, let's analyze the situation. We have a hand-pumped water gun that is held level above the ground, and when it is fired, the water stream hits the ground at a certain horizontal distance. We are asked to find the gauge pressure of the water gun's reservoir at the instant it is fired.
To solve this problem, we can use Bernoulli's equation, which relates the pressure, velocity, and height of a fluid. In this case, we can assume that the speed of the water in the reservoir is zero and that the water flow is steady.
Let's apply Bernoulli's equation:
P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂
Where:
P₁ and P₂ are the initial and final pressures (gauge pressure in this case),
ρ is the density of water,
v₁ and v₂ are the initial and final velocities of the water,
g is the acceleration due to gravity, and
h₁ and h₂ are the initial and final heights.
In this case, we can assume that the height of the reservoir (h₁) is constant, so h₁ - h₂ = 0. The velocity of the water in the reservoir (v₁) is zero, so we can simplify the equation:
P₁ + 0 + 0 = P₂ + (1/2)ρv₂² + 0
This simplifies to:
P₁ = P₂ + (1/2)ρv₂²
Now, we need to find the final velocity of the water (v₂). We can use the horizontal distance the water stream travels (6.9 m) and the equation of motion for horizontal velocity:
v = d/t
Where v is the velocity, d is the distance, and t is the time. If we assume that the water stream hits the ground horizontally and we neglect air resistance, the velocity of the water stream (v₂) is equal to the horizontal velocity.
So, v₂ = 6.9 m/s
Now, we can plug this value into the equation to find P₁:
P₁ = P₂ + (1/2)ρv₂²
We are given that the water gun is held at a height of 0.85 m above the ground, so we can find the value of P₂ using the following equation:
P₂ = Po + ρgh₂
Where Po is atmospheric pressure.
Finally, we can substitute the values and solve the equation to find P₁, the gauge pressure of the water gun's reservoir at the instant it is fired.
I hope this explanation helps you understand how to approach this problem.
Not to worry! Let's break down the problem step-by-step and work through it together.
Step 1: Start by identifying the given information in the problem:
- The height of the water gun above the ground is 0.85 m.
- The horizontal distance traveled by the water stream is 6.9 m.
- The speed of the water in the reservoir is zero.
- The water flow is steady.
- Air resistance and the height difference between the reservoir and the muzzle can be ignored.
Step 2: Draw a diagram to visualize the problem. It can be helpful to draw a horizontal line to represent the ground and a vertical line to represent the height of the water gun above the ground.
/ | <-- Height of the water gun (0.85 m)
/
/
/
/ <-- Water stream hits the ground here (6.9 m)
/
/
/
/
-----------/
Ground
Step 3: Identify the relevant equation(s) to solve the problem.
In this case, we can use the equation for pressure due to height difference:
P1 + ρgh1 = P2 + ρgh2
Where:
P1 = initial pressure (gauge pressure of the water gun's reservoir)
P2 = final pressure (at the height of the water stream hitting the ground)
ρ = density of water (constant)
g = acceleration due to gravity (constant)
h1 = initial height (height of the water gun above the ground)
h2 = final height (height of the water stream hitting the ground)
Step 4: Plug in the known values into the equation and solve for the initial pressure (P1).
To solve this equation, we need to find the difference in height between the initial and final positions. We can subtract h2 from h1:
Δh = h2 - h1
Plugging in the values:
Δh = 0 - 0.85 = -0.85 m
Now we can calculate the gauge pressure (P1):
P1 = P2 + ρgΔh
Step 5: Calculate the density of water (ρ) and the acceleration due to gravity (g).
The density of water (ρ) is approximately 1000 kg/m³, and the acceleration due to gravity (g) is approximately 9.8 m/s².
Substitute these values into the equation and calculate the initial pressure (P1).
P1 = P2 + (1000 kg/m³) × (9.8 m/s²) × (-0.85 m)
Step 6: Calculate the value of P1.
P1 = P2 + (-8330 N/m²)
Step 7: Calculate the final pressure (P2).
To calculate the final pressure, we need to consider that the speed of the water in the reservoir is zero. This means that the pressure in the reservoir is equal to atmospheric pressure (approx. 101,325 N/m²).
P2 = 101,325 N/m²
Step 8: Substitute the value of P2 into the equation from Step 6 and calculate P1.
P1 = 101,325 N/m² + (-8330 N/m²)
P1 ≈ 93,995 N/m²
Therefore, the gauge pressure of the water gun's reservoir at the instant it is fired is approximately 93,995 N/m².