Question

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.

Answer 1

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
$$\begin{gathered} v=\sqrt{2g(\mathrm{H}-h)} \\ \mathrm{Also},\mathrm{let}t\mathrm{be}\mathrm{the}\mathrm{time}\mathrm{of}\mathrm{flight}. \\ \mathrm{Now}, \\ t=\sqrt{\frac{2h}{g}} \\ \mathrm{Let}x\mathrm{be}\mathrm{the}\mathrm{maximum}\mathrm{horizontal}\mathrm{distance}. \\ \therefore x=v\times t \\ =\sqrt{2g(\mathrm{H}-h)}\times \sqrt{\frac{2h}{g}} \\ =\sqrt{4(\mathrm{H}h-h^2)} \end{gathered}$$
For x to be maximum,
$$\begin{gathered} \frac{d}{dh}\left(\mathrm{H}h-h^2\right)=0 \\ \Rightarrow 0=\mathrm{H}-2h \\ \Rightarrow h=\frac{\mathrm{H}}{2} \end{gathered}$$