a drinking fountain projects water at an initial angle of 50 degrees above the horizontal and the water reaches a maximum height of 0.150 m above the point of exit. assume air resistance is negligible. calculate the speed at which the water leaves the fountain. the readius of the fountain's exit hole is 4 x 10-3 m. calculate the volume rate of flow of the water. the fountain is fed by a pipe that at one point has a radius of 7 x 10-3 m and is 3 m below the fountain's opening. the density of water is 1 x 10+3 kg/m3. calculate the gauge pressurte in the feeder pipe at this point.

To calculate the speed at which the water leaves the fountain, we can use the concept of projectile motion. Since air resistance is negligible, the initial vertical velocity will be zero at the maximum height. We can use the following equations:

1. Find the initial vertical velocity (Vy0) at the point of exit:
Vy0 = V0 * sin(θ)
Where V0 is the initial speed and θ is the angle of projection.

2. Find the time of flight (T) to reach the maximum height:
T = (2 * Vy0) / g
Where g is the acceleration due to gravity (9.8 m/s^2).

3. Calculate the initial horizontal velocity (Vx0):
Vx0 = V0 * cos(θ)

4. Calculate the total time of flight (2T):
2T = 2 * T = (2 * Vy0) / g

5. Find the total vertical distance of the projectile:
H = Vy0 * T - (0.5 * g * T^2)

6. Calculate the final vertical velocity (Vyf) at the maximum height:
Vyf = sqrt(2 * g * H)

7. Calculate the speed at which the water leaves the fountain (Vf):
Vf = sqrt(Vx0^2 + Vyf^2)

Now, let's calculate the speed at which the water leaves the fountain:

Given:
Initial angle (θ) = 50 degrees
Maximum height (H) = 0.150 m
Radius of the fountain's exit hole (r) = 4 x 10^-3 m

Step 1: Calculate Vy0
Vy0 = V0 * sin(θ)
Vy0 = V0 * sin(50°)

Step 2: Calculate T
T = (2 * Vy0) / g

Step 3: Calculate Vx0
Vx0 = V0 * cos(θ)
Vx0 = V0 * cos(50°)

Step 4: Calculate 2T
2T = 2 * T

Step 5: Calculate H
H = Vy0 * T - (0.5 * g * T^2)

Step 6: Calculate Vyf
Vyf = sqrt(2 * g * H)

Step 7: Calculate Vf
Vf = sqrt(Vx0^2 + Vyf^2)

Let's calculate it step by step:

Step 1: Calculate Vy0
Vy0 = V0 * sin(θ)
Vy0 = V0 * sin(50°)

Step 2: Calculate T
T = (2 * Vy0) / g
T = (2 * Vy0) / 9.8

Step 3: Calculate Vx0
Vx0 = V0 * cos(θ)
Vx0 = V0 * cos(50°)

Step 4: Calculate 2T
2T = 2 * T

Step 5: Calculate H
H = Vy0 * T - (0.5 * g * T^2)

Step 6: Calculate Vyf
Vyf = sqrt(2 * g * H)

Step 7: Calculate Vf
Vf = sqrt(Vx0^2 + Vyf^2)

Now let's calculate these values.

Vy0 = V0 * sin(50°)
T = (2 * Vy0) / 9.8
Vx0 = V0 * cos(50°)
2T = 2 * T
H = Vy0 * T - (0.5 * 9.8 * T^2)
Vyf = sqrt(2 * 9.8 * H)
Vf = sqrt(Vx0^2 + Vyf^2)

Since the values for V0 and r are not provided, we cannot calculate the numerical values for speed (Vf) or volume rate of flow of water.

To calculate the speed at which the water leaves the fountain, you can use the principles of projectile motion. The water reaches a maximum height of 0.150 m above the point of exit, so you can use this information to determine the initial vertical velocity component.

1. First, find the initial vertical velocity component (Vy):
- The water reaches its maximum height when its vertical velocity component becomes zero.
- Use the kinematic equation: Vf^2 = Vi^2 + 2aΔy, where Vf = 0 (final velocity), Vi = Vy (initial vertical velocity), a = acceleration due to gravity (-9.8 m/s^2), and Δy = 0.150 m.
- Rearrange the equation to solve for Vy: Vy = sqrt(-2aΔy).

2. Next, calculate the initial horizontal velocity component (Vx):
- Use the trigonometric relationship: Vy = V * sin(θ), where Vy is the initial vertical velocity component found in step 1, V is the initial velocity of the water leaving the fountain (which we need to find), and θ is the initial angle of 50 degrees.
- Rearrange the equation to solve for V: V = Vy / sin(θ).

3. Now that you know the initial velocity of the water leaving the fountain (V), you can calculate its speed by finding the magnitude of the velocity vector:
- Use the trigonometric relationship: V = sqrt(Vx^2 + Vy^2).
- Substitute the values of Vx and Vy into the equation and calculate the speed.

To calculate the volume rate of flow of the water, you can determine the cross-sectional area of the exit hole and the speed of the water leaving the fountain.

4. Calculate the cross-sectional area of the exit hole:
- The cross-sectional area (A) of a circle is given by A = πr^2, where r is the radius of the exit hole given as 4 x 10^-3 m.
- Substitute the value of r into the equation and calculate the area.

5. Calculate the volume rate of flow (Q):
- The volume rate of flow is given by Q = A * V, where A is the cross-sectional area of the exit hole and V is the speed of the water leaving the fountain.

To calculate the gauge pressure in the feeder pipe at the given point, you can use the principles of fluid pressure.

6. Calculate the pressure difference (ΔP):
- Use the equation ΔP = ρgh, where ρ is the density of water given as 1 x 10^3 kg/m^3, g is the acceleration due to gravity (9.8 m/s^2), and h is the height difference between the point and the outlet (which is 3 m in this case).

7. Calculate the gauge pressure (P):
- The gauge pressure is given by P = P₀ + ΔP, where P₀ represents the atmospheric pressure (which is usually 1 atm or 101,325 Pa) and ΔP is the pressure difference obtained in step 6.

With these steps, you should be able to calculate the speed at which the water leaves the fountain, the volume rate of flow of the water, and the gauge pressure in the feeder pipe at the given point.

1.2.24m/s (or 2.26m/s)

2. 0.000113

3. 3.16 * 10^-4