Question

You hold a hose at waist height and spray water horizontally with it. The hose nozzle has a diameter of 1.80 cm, and the water splashes on the ground a distance of 0.950 cm horizontally from the nozzle. If you constrict the nozzle to a diameter of 0.750 cm, how far from the nozzle, horizontally, will the water travel before it hits the ground? (Ignore air resistance.)

Short answer

Answer

The horizontal distance travelled by water before it hits the ground is, 5.47m .

Step-by-step solution

Step-by-Step Solution Step 1: Identification of the given data

The given data can be listed below as,

• The hose nozzle diameter is, d₂ = 1.80 cm .
• The horizontal distance of water splashes is, x₂ = 0.950 m .
• The constrict nozzle diameter is, d₁ = 0.750cm .

Significance of continuity equation

The amount of water flow (volume flow rate) at any point in unit time will be equal to all the points in the fluid. The continuity equation is used to calculate the area, and velocity at the required point.

Determination of the horizontal distance travel by water before it hits the ground

If the nozzle or hose constrict, it means area is decreased, and velocity is increased. Applying continuity equation on the water flow as,

$\begin{array}{c}{A}_1{v}_1=A_2{v}_2\\ \frac{v_1}{v_2}=\frac{A_2}{A_1}\\ \frac{v_1}{v_2}=\frac{\frac{\pi }{4}{d}_2^2}{\frac{\pi }{4}{d}_1^2}\\ \frac{v_1}{v_2}=\frac{d_2^2}{d_1^2}......\left(i\right)\end{array}$

Here, v₁ and v₂ are the velocity of water at constrict nozzle and hose nozzle, d₁ and d₂ are the diameter at constrict nozzle and hose nozzle.

The relation of distance, time and velocity is expressed as,

x = vt

Here, x is the horizontal distance travel by water; t is the time, and v is the velocity of water. For the intel and outlet condition is expressed as,

$\begin{array}{c}{x}_1=v_{1}t\\ x_2=v_{2}t\\ \frac{x_1}{x_2}=\frac{v_{1}t}{v_{2}t}\\ =\frac{v_1}{v_2}\end{array}$

Here x₁ is the horizontal distance travel by water, and x₂ is thehorizontal distance of water splashes.

Take the value of $\frac{v_1}{v_2}$from the equation (I) and substitute in the above equation.

$\begin{array}{c}\frac{x_1}{x_2}=\frac{d_2^2}{d_1^2}\\ x_1=x_2\frac{d_2^2}{d_1^2}\end{array}$

Substitute 0.0075m for d₁ , 0.018m for d₂ and 0.95 for x₂ in the above equation.

$\begin{array}{c}{x}_1=0.95\text{\hspace{0.17em}m}\times {\left(\frac{0.018\text{\hspace{0.17em}m}}{0.0075\text{\hspace{0.17em}m}}\right)}^2\\ =5.47\text{\hspace{0.17em}m}\end{array}$

Hence, the horizontal distance travelled by water before it hits the ground is,5.47m .