The water is filled upto height of 12 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.

Answer

To solve the problem of finding the depth \( h \) below the water level at which a hole should be made in a tank to maximize the range of the emerging water stream, we can follow these steps:

### Step 1: Understand the problem
We have a tank filled with water to a height of 12 m. A hole is made at a depth \( h \) below the water level. We need to find the value of \( h \) that maximizes the horizontal distance (range) that the water stream travels when it exits the hole.

### Step 2: Determine the velocity of the water exiting the hole
The velocity \( v \) of the water emerging from the hole can be determined using Torricelli’s theorem, which states that the speed of efflux of a fluid under the force of gravity through an orifice is given by:
\[ v = \sqrt{2gH'} \]
where \( H' \) is the height of the water column above the hole. Since the total height of the water is 12 m and the hole is at depth \( h \), we have:
\[ H' = 12 - h \]
Thus, the velocity becomes:
\[ v = \sqrt{2g(12 - h)} \]

### Step 3: Determine the time of flight
The time \( t \) it takes for the water to fall from the hole to the ground can be calculated using the equation of motion:
\[ s = ut + \frac{1}{2}gt^2 \]
Here, \( s = h \) (the distance the water falls), \( u = 0 \) (initial vertical velocity), and \( g \) is the acceleration due to gravity. Therefore:
\[ h = 0 \cdot t + \frac{1}{2}gt^2 \implies h = \frac{1}{2}gt^2 \]
From this, we can solve for \( t \):
\[ t = \sqrt{\frac{2h}{g}} \]

### Step 4: Calculate the range
The horizontal range \( R \) of the water stream can be expressed as:
\[ R = v \cdot t \]
Substituting the expressions for \( v \) and \( t \):
\[ R = \sqrt{2g(12 - h)} \cdot \sqrt{\frac{2h}{g}} = 2\sqrt{h(12 - h)} \]

### Step 5: Maximize the range
To find the value of \( h \) that maximizes \( R \), we can differentiate \( R \) with respect to \( h \) and set the derivative equal to zero:
\[ \frac{dR}{dh} = 2 \cdot \frac{1}{2\sqrt{h(12 - h)}} \cdot (12 - 2h) = 0 \]
Setting the term \( (12 - 2h) = 0 \) gives:
\[ 12 - 2h = 0 \implies h = 6 \]

### Conclusion
The value of \( h \) for which the emerging stream of water strikes the ground at the maximum range is:
\[ \boxed{6 \text{ m}} \]