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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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