Water level is maintained in  cylindrical vessel upto a fixed height 'H'. The vessel is kept on a horizontal plane. At what height above the bottom should a hole be made in the vessel so that the water stream coming out of the hole strikes the horizontal plane at the greatest distance from the vessel.

let hole be made at height 'h' from the ground

so, pressure at the level wher hole is made is given by

P = (H-h)g

and F= PA = (H-h)gA

acceleration of water = a = F/m = (H-h)gA/m

Now, time taken by the stream to fall on the ground from height 'h' is

t = 2gh

Therefore, Horizontal distance travelled in time 't' is

x = ut + (1/2)at²

or, x = (1/2) [ (H-h)gA/m] [2gh]

or, x = (H-h)g²Ah/m

so, for x = maximum

dx/dh = 0

dx/dh = -g²Ah/m +(H-h)g²A/m

or (H-h)g²A/m = g²Ah/m

or (H-h) = h

or h = H/2 is the reqiured solution