A cylindrical vessel contains a liquid up to a height h. A hole is made on the wall of the vessel. Water comes out through the hole and falls at a distance x from the base of the vessel. At what depth should the hole be created so that x becomes maximum?

Explanation

Suppose a hole is made at a depth y from the free surface of the liquid and the horizontal velocity of effux of water through the hole is v.
`therefore" "v=sqrt(2gy)`
Suppose the liquid takes t second to fall at a distance x from the base of the vessel. Considering the vertical motion of the ejected liquid, we get
`h-y=1/2g t^(2)or,t^(2)=(2(h-y))/g`

Considering the horizontal motion of the ejected liquid, we get
`x=vt or,x=sqrt(2gy)t`
or, `x^(2)=2gyt^(2)=2gy*(2(h-y))/g=4y(h-y)" "...(1)`
Now, for the maximum value of x, `x^(2)` is also maximum and hence `d/(dy)(x^(2))=0`.
So, differentiating equation (1) with respect to y, we get
`d/(dy)(x^(2))=4(h-y)+4y(-1)=4(h-2y)`
`therefore" "h-2y=0or,y=h/2`
Hence, for the maximum range of the water coming out of the hole, the should be made at a depth of `h/2`.