Question 14.7: Watering a Garden A gardener uses a water hose to fill a 30....
Watering a Garden
A gardener uses a water hose to fill a 30.0-L bucket. The gardener notes that it takes 1.00 min to fill the bucket. A nozzle with an opening of cross-sectional area 0.500 cm² is then attached to the hose. The nozzle is held so that water is projected horizontally from a point 1.00 m above the ground. Over what horizontal distance can the water be projected?
Conceptualize Imagine any past experience you have with projecting water from a horizontal hose or a pipe using either your thumb or a nozzle, which can be attached to the end of the hose. The faster the water is traveling as it leaves the hose, the farther it will land on the ground from the end of the hose.
Categorize Once the water leaves the hose, it is in free fall. Therefore, we categorize a given element of the water as a projectile. The element is modeled as a particle under constant acceleration (due to gravity) in the vertical direction and a particle under constant velocity in the horizontal direction. The horizontal distance over which the element is projected depends on the speed with which it is projected. This example involves a change in area for the pipe, so we also categorize it as one in which we use the continuity equation for fluids.
Analyze
Express the volume flow rate I_V in terms of area and speed of the water in the hose (we will discuss the origin for this notation in Section 14.7):
I_V=A_1 v_1Solve for the speed of the water in the hose:
v_1=\frac{I_V}{A_1}We have labeled this speed v_1 because we identify point 1 within the hose. We identify point 2 in the air just outside the nozzle. We must find the speed v_2=v_{x i} with which the water exits the nozzle \left(v_2\right) and begins its projectile motion \left(v_{x i}\right). The subscript i anticipates that it will be the initial velocity component of the water projected from the hose, and the subscript x indicates that the initial velocity vector of the projected water is horizontal.
Solve the continuity equation for fluids for v_2 :
(1) v_2=v_{x i}=\frac{A_1}{A_2} v_1=\frac{A_1}{A_2}\left(\frac{I_V}{A_1}\right)=\frac{I_V}{A_2}
We now shift our thinking away from fluids and to projectile motion. In the vertical direction, an element of the water starts from rest and falls through a vertical distance of 1.00 m.
Write Equation 2.16 for the vertical position of an element of water, modeled as a particle under constant acceleration:
x_f=x_i+v_{x i} t+\frac{1}{2} a_x t^2 \quad\left(\text { for constant } a_x\right) (2.16)
y_f=y_i+v_{y i} t-\frac{1}{2} g t^2Identify the origin as the initial position of the water as it leaves the hose, and recognize that the water begins with a vertical velocity component of zero. Solve for the time at which the water reaches the ground:
(2) y_f=0+0-\frac{1}{2} g t^2 \rightarrow t=\sqrt{\frac{-2 y_f}{g}}
Use Equation 2.7 to find the horizontal position of the element at this time, modeled as a particle under constant velocity:
x_f=x_i+v_x t \quad\left(\text { for constant } v_x\right) (2.7)
x_f=x_i+v_{x i} t=0+v_2 t=v_2 tSubstitute from Equations (1) and (2):
x_f=\frac{I_V}{A_2} \sqrt{\frac{-2 y_f}{g}}Substitute numerical values:
x_f=\frac{30.0 L / \min}{0.500 cm^2} \sqrt{\frac{-2(-1.00 m)}{9.80 m / s^2}}\left(\frac{10^3 cm^3}{1 L}\right)\left(\frac{1 \min}{60 s}\right)=452 cm=4.52 mFinalize The time interval for the element of water to fall to the ground is unchanged if the projection speed is changed because the projection is horizontal. Increasing the projection speed results in the water hitting the ground farther from the end of the hose, but requires the same time interval to strike the ground.