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Discussion Response Post to: (FLUID MECHANICS)!

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CyBorG (1574)

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The solution is as follows....

Let the small hole be the origin.

Then the eqn of inclined plane is y+h=x/(root3)

or x=(root3)(y+h)........(1)

For the water coming out x=vt where v=root(2g(10-h))

and -y=gt²/2

So y=x²/4(h-10)

So dy/dx=x/2(h-10).....(2) which is the slope of the tangent at the point where water meets the inclined plane.

From (1) slope of the normal is 1/(root3)

So x/2(h-10)= -(root3)

or x=2(root3)(10-h)

So y=3(h-10)

Substtute this in (1) you will get h=50/6=8.33m

-ADARSH NITK Surathkal

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