Water level is maintained in a cylindrical vessel up to 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.
Solution
It is given that H is the height of the cylindrical vessel.
Now, let h be the height of the hole from the surface of the tank.
The velocity of water \[\left( v \right)\] is given by
\[v = \sqrt{2g(H - h)}\]
Also, let t be the time of flight.
Now,
\[t = \sqrt{\frac{2h}{g}}\]
Let x be the maximum horizontal distance .
\[ \therefore x = v \times t\]
\[ = \sqrt{2g(H - h)} \times \sqrt{\frac{2h}{g}}\]
\[ = \sqrt{4(Hh - h^2 )}\]
For x to be maximum,
\[\frac{d}{dh}\left( Hh - h^2 \right) = 0\]
\[ \Rightarrow 0 = H - 2h\]
\[ \Rightarrow h = \frac{H}{2}\]
