Fluids (AP Physics)

A drinking fountain projects water at an initial angle of 50 degrees above the horizontal, and the water reaches a maximum height of 0.150m above the point of exit. Assume air resistance is negligible. a. Calculate the speed at which the water leaves the fountain. 

b. The radius of the fountain’s exit hole is 4.00 X 10 –3m (10 to the –3). Calculate the volume rate of flow of water. 

c. The fountain is fed by a pipe that at one point has a radius of 7.00 X 10 –3m (10 to the –3) and is 3.00 m below the fountain’s opening . The density of water is 1.00 X 10 3 (ten to the third) kg/m 3 (M to the 3rd power. Calculate the gauge pressure in the feeder pipe at this point. Please provide the work, step by step that arrives at the answers. Thank You!


a) The vertical component of the speed V0y can be deduced from

$0 = V0y^2 -2 g h$, where h is the maximum height

$V0y = \sqrt{2*g*h}= \sqrt{2*9.81*0.150} =1.715 m/s$

Since $V0y = V0*sin(\alpha)$ , where $V0$ is the initial speed

$V0 = V0y/sin(\alpha) = 1.715/sin(50) =2.239 m/s$

b) The volume rate is

$Volume rate = \pi*r^2*V0 = \pi*(4*10^{-3})^2*2.239 = 1.125*10^{-4} m^2/sec$

c) The speed of water in the pipe comes from (the dynamic pressure $\rho*v^2/2$ is also a force over a surface)

$V0*S1 = V*S2$

$V = V0*S1/S2 =V0*(r1/r2)^2 = 2.239*(4/7)^2 =0.731 m/s$

For the sum of the pressures one has

$P_{atm} +\rho*V0^2/2 = P+\rho*V^2/2+\rho*g*H$

where H is the deepness of the pipe below the fountain opening.

$P = P_{atm}+\rho/2*(V0^2-V^2)-\rho*g*H =$

$=P_{atm}+1000/2*(2.239^2-0.731^2)-1000*9.81*3 =$

$=P_{atm} -28310.12 N/m^2 =10^5 -28310.12 =71689.88 N/m^2$