A fountain whose opening is \(2 \mathrm{~cm}\) in diameter sprays water vertically into the air with a velocity of \(5 \mathrm{~m} / \mathrm{s}\). You may take \(g=10 \mathrm{~m} / \mathrm{s}^2\). a) What is the maximum height that the water goes into the air? b) What is the volume flow rate? c) If the fountain is fed by a feeder pipe with diameter \(3 \mathrm{~cm}\), what is the absolute pressure in the feeder pipe \(1.5 \mathrm{~m}\) below the surface of the water? d) If the designer of the fountain wants the water to reach a height of \(4 \mathrm{~m}\) what should he or she do without altering the volume flow rate?

To find the maximum height that the water goes into the air, we will use the following kinematic equation of motion: v^2 = u^2 + 2 * a * h Here, v = 0 m/s (at maximum height, the velocity is zero), u = 5 m/s (initial velocity), a = -g = -10 m/s^2 (acceleration due to gravity), and h (height) is what we need to find. Let's rearrange the equation to solve for h: 0 = 5^2 - 2 * 10 * h Now, let's solve for h: h = (5^2) / (2 * 10) h = 25 / 20 h = 1.25 m So, the maximum height that the water goes into the air is 1.25 meters.

The volume flow rate can be calculated using the formula: Q = A * v Here, A is the cross-sectional area of the fountain opening, and v is the velocity of the water being sprayed. First, we need to find the cross-sectional area of the opening. It is given that the diameter of the opening is 2 cm. The area can be calculated using the formula: A = π * (d/2)^2 A = π * (2/2)^2 A = π * 1^2 A = π cm^2 Now, we can calculate the volume flow rate: Q = A * v Q = π * 5 Q = 5π m³/s Thus, the volume flow rate is 5π cubic meters per second.

To find the absolute pressure in the feeder pipe, we will use the Bernoulli's equation. Bernoulli's equation states that for an incompressible flow, the total mechanical energy per unit volume is constant: P_1 + 0.5 * ρ * v_1^2 + ρ * g * h_1 = P_2 + 0.5 * ρ * v_2^2 + ρ * g * h_2 Here, P_1 and P_2 are the pressures at two points, v_1 and v_2 are the velocities, h_1 and h_2 are the heights above the reference level, and ρ is the density of the fluid (water). Let's assume that the density of water is 1000 kg/m³. The feeder pipe has a diameter of 3 cm and is 1.5 m below the surface. First, let's calculate the cross-sectional area of the feeder pipe: A_f = π * (d/2)^2 A_f = π * (3/2)^2 A_f = (9/4)π cm^2 Now, we need to find the velocity of the water in the feeder pipe. Since the volume flow rate is constant, we can use: Q = A * v Q = A_f * v_f Solving for v_f, we get: v_f = Q / A_f v_f = 5π / (9/4)π v_f = (20/9) m/s Now, let's use the Bernoulli equation at the surface (point 1) and the feeder pipe (point 2): P_1 = atmospheric pressure = P_atm P_2 = absolute pressure in the feeder pipe (what we need to find) v_1 = 0 m/s (surface) v_2 = (20/9) m/s (feeder pipe) h_1 = 0 m h_2 = 1.5 m Plugging the values into the Bernoulli equation, we get: P_atm + 0.5 * 1000 * 0^2 + 1000 * 10 * 0 = P_2 + 0.5 * 1000 * (20/9)^2 + 1000 * 10 * 1.5 Now, let's solve for P_2: P_2 = P_atm - 0.5 * 1000 * (20/9)^2 - 1000 * 10 * 1.5 So, the absolute pressure in the feeder pipe 1.5 meters below the surface of the water is P_2.

The designer can increase the water height without altering the volume flow rate by either increasing the velocity of the water being sprayed or reducing the diameter of the fountain opening. By doing so, the designer can maintain the same volume flow rate while increasing the kinetic energy, which results in the water reaching a higher height.