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Question

A rectangular tank of height 10m filled with water, is placed near the bottom of a plane inclined at an angle 30 with horizontal. At height h from bottom a small hole is made (as shown in figure) such that the stream coming out from hole, strikes the inclined plane normally. Calculate h
220569_0e63e84c8f4a4d238229cc8bcfae122c.png

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Solution

v=2g(10h)
Component of its velocity parallel to the plane is vcos30.
Let the stream strikes the plane after time t. Then
0=vcos30gsin30t
t=vcos30g
Further x=vt=v2cos30g
=3y
or v2cos30g=3(h12gt2)
3v2g=3(hg2v2cos230g2)
or v2g=h32v2g
52v2g=h
or 5(10h)=h
h=8.33m

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