Question

A horizontal pipe of diameter D = 3.0 cm passes through a reservoir dam at depth d = 5.0 m. A plug secures the pipe opening. The water density is p = 1.0 x10^3 kg/m3. If the pipe is h = 10.0 m above the ground, how FAR will the water travel horizontally from the pipe exit when hitting the ground? Ignore air friction.

Answer 1

The velocity of the water when it leaves the hole is given by:

$v=\sqrt[]{2gd}$

where g is the gravity and d is the depth of the pipe. Then in this case we have:

$v=\sqrt[]{2(9.8)(5)}=98$

hence the velocity of the water when it leaves the pipe is 98 m/s.

Now that we know that we can calculate the horizontal distance using the relation:

$v=\frac{d}{t}$

where d is the distance and t is the time. To use this relation we need the time it takes the water to hit the ground, to find it we need to remember that in the vertical direction the time can be obtained from the formula:

$y-y_0=v_0t-\frac{1}{2}gt^2$

(we can use this formula since we can think that the vertical motion follows a free fall).

From this (assuming the ground is y=0) we have:

$\begin{gathered} 0-10=(0)t-\frac{1}{2}(9.8)t^2 \\ -10=-4.9t^2 \\ t^2=\frac{-10}{-4.9} \\ t=\sqrt[]{\frac{10}{4.9}} \\ t=1.429 \end{gathered}$

Then it takes the water 1.429 (rounded to three decimals places) seconds to hit the ground. Once we know this we can use the equation of velocity to find the distance from the pipe:

$\begin{gathered} 98=\frac{d}{1.429} \\ d=(1.429)(98)=140.042 \end{gathered}$

Therefore the water will hit the floor 140.042 meters from the pipe.