A tank open on top has a depth h and is filled with water. A hole is made three-quarter from the ground and another hole is made half the depth. What is the difference in the range of water ejected through these holes?

1 Answer
Mr. Mike
Mar 31, 2018

The difference in range will be the height of the level of water in the tank times #(1-sqrt(3)/2)#.

Explanation:

You really want to use Torricelli's Law here.

#v=sqrt(2g(h-x))#

where

#v# = the horizontal velocity of the stream of water,

#g# = the acceleration of gravity, and

#h# = the vertical distance from the surface of the water to the ground, and

#x# = is the vertical height of the hole from the ground (see diagram).

Original Image From Algebra By Hand(TM)

The range of the stream is the horizontal distance the stream lands away from the tank. For the halfway hole, I have labeled this #R_1#, and for the 3/4 hole, I have labeled this #R_2#.

How long will it take a water droplet to reach the ground once it has left the hole? Since it's vertical velocity component is controlled by gravity, the time it takes to land on the ground, #t#, will be

#t=sqrt((2x)/g)#.

For the halfway hole, #x=h/2#, and for the 3/4 hole, #x=(3h)/4#.

Now we need to figure out the horizontal distance that the drop will travel in that time. It will simply be its horizontal velocity times the time.

#R=vt#

We can find #v# from Torricelli's Law and #t# from above, so we will have

#R=sqrt(2g(h-x))sqrt((2x)/g)=2sqrt((h-x)x)#

The problem statement asks us to determine

#R_1-R_2=2sqrt((h-h/2)h/2)-2sqrt((h-(3h)/4)(3h)/4#

#=2h[sqrt((1-1/2)1/2)-sqrt((1-(3)/4)3/4)]#

#=2h(1/2-sqrt(3)/4)=h(1-sqrt(3)/2)#

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