IAS Civil Exam, UPPSC Entrance Preparation, Civil Services Tests Online,Disscussion Forum

CyBorG

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