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Question

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


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Solution


Speed of efflux, v0=2g(10h)
Component of its velocity parallel to the plane is v0cos30

Let the stream strike the plane after time t. Then, along the plane,
0=v0cos30gsin30t
[Since the stream strikes perpendicular to the plane]
t=v0cot30g

Along horizontal direction,
x=v0t=v20cot30g(1)
Also, from ΔOAB,
x=3(hy)=3(h12gt2)

3v20g=3(h12×g×v20cot230g2)
v20g=h3v202g
h=5v202g
h=52g×2g(10h)h=5(10h)
h=8.33 m

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