Animated Solution for Physics - Properties of Solids and Liquids: The water is filled upto height of $12\text{ m}$ in a tank having vertical sidewalls. A hole is made in one of the walls at a depth $h$ below the water level. The value of $h$ for which the emerging stream of water strikes the ground at the maximum range is ...... m.

A Leaky Tank

Imagine a large tank resting on the ground, filled with water up to a total height of $H=12\text{ m}$. Now, suppose we puncture a small hole in the vertical sidewall at a depth $h$ from the free surface of the water.

Water will immediately begin to spurt out, forming a parabolic trajectory before hitting the ground. The horizontal distance from the base of the tank to the point where the water lands is called the range, $R$. Our goal is to find the exact depth $h$ that maximizes this range.

The Speed of Efflux

To find the range, we first need to know how fast the water is leaving the tank. According to Torricelli's Law, the velocity of efflux $v$ of a fluid from a small hole is the same as the velocity an object would acquire falling freely from rest through a height equal to the depth of the hole.

Mathematically, this is derived from Bernoulli's principle, giving us: where $g$ is the acceleration due to gravity. Notice that this horizontal velocity depends only on the depth $h$ of the hole from the top surface.

Time and Range

Once the water leaves the hole, it acts like a horizontal projectile. The time it takes to hit the ground depends entirely on the vertical distance it has to fall. Since the total height of the water is $12\text{ m}$ and the hole is at a depth $h$, the height of the hole from the ground is $(12-h)$.

Using the second equation of motion for vertical free fall ($s=ut+\frac{1}{2}a{t}^2$), with initial vertical velocity $u=0$:

The horizontal range $R$ is simply the constant horizontal velocity multiplied by the time of flight:

Maximizing the Range

Let's simplify the expression for the range. Notice how beautifully the acceleration due to gravity $g$ cancels out:

To find the maximum range, we need to find the value of $h$ that maximizes this expression. To make the calculus easier, instead of differentiating the square root, we can just maximize the square of the range, $R^2$. Let's call it $y$:

Now, we differentiate $y$ with respect to $h$ and set it to zero to find the critical point:

The math tells us that the hole must be exactly $6\text{ m}$ from the top. Since the total height is $12\text{ m}$, this means the hole must be placed exactly at the midpoint of the water column!

This is a profound and standard result in fluid mechanics: For any tank of height $H$ resting on the ground, the maximum horizontal range is achieved when the hole is at $h=\frac{H}{2}$. Furthermore, if you substitute this back into the range equation, you'll find that the maximum range itself is exactly equal to the total height of the water, $R_{\text{max}}=H$.