Question

Water stands at a depth $H$ in a tank whose side wails are vertical. A hole is made on one of the walls at a depth $h$ below the water surface. Find at what distance from the foot of the wall does the emerging stream of water strike the floor?

Answer 1

The velocity is given as

$v=\sqrt{2gh}$

Let the hole was made at height$h$below the water level

Then

$H-h=\frac{1}{2}g{t}^2$

By using the equation

$v=\frac{d}{t}$

$R=v\times t$

By substituting the time from the above equation we get

$R=\sqrt{2gh}\times {\left(\frac{2\left(H-h\right)}{g}\right)}^{\frac{1}{2}}$

On solving the above equation

$R=2\sqrt{h\left(H-h\right)}$

The maximum range can be found by

$\frac{dR}{dh}=0$

Hence the maximum range is$H=2h$