A rectangular tank is placed on a horizontal ground and is filled with water to a height H above the base. A small hole is made on one vertical side at a depth D below the level of the water in the tank. The distance x from the bottom of the tank at which the water jet from the tank will hit the ground is

Options

A. ` 2[D(H-D)]]^(1//2)`
B. ` 2 (gD)^(1/2)`
C. `2[D(H+D)]]^(1//2)`
D. `1/2 (DH)^(1/2)`

Answer

To solve the problem of determining the distance \( x \) from the bottom of the tank at which the water jet will hit the ground, we can follow these steps:

### Step 1: Determine the height of the water above the hole
The height of the water above the hole is given by:
\[ h_{\text{above}} = H - D \]
where \( H \) is the total height of the water in the tank and \( D \) is the depth of the hole below the water level.

### Step 2: Calculate the efflux speed of the water
Using Torricelli's theorem, the speed \( v \) of the water exiting the hole can be calculated using the formula:
\[ v = \sqrt{2g h_{\text{above}}} = \sqrt{2g (H - D)} \]
where \( g \) is the acceleration due to gravity.

### Step 3: Determine the time of flight of the water jet
The time \( t \) it takes for the water to fall from the height \( H - D \) to the ground can be calculated using the second equation of motion:
\[ H - D = \frac{1}{2} g t^2 \]
Rearranging this gives:
\[ t^2 = \frac{2(H - D)}{g} \]
Taking the square root, we find:
\[ t = \sqrt{\frac{2(H - D)}{g}} \]

### Step 4: Calculate the horizontal distance traveled by the water jet
The horizontal distance \( x \) that the water travels while falling can be calculated using the formula:
\[ x = v \cdot t \]
Substituting the expressions for \( v \) and \( t \):
\[ x = \sqrt{2g(H - D)} \cdot \sqrt{\frac{2(H - D)}{g}} \]
Simplifying this expression:
\[ x = \sqrt{2g(H - D)} \cdot \sqrt{\frac{2(H - D)}{g}} = 2(H - D) \]

### Final Result
Thus, the distance \( x \) from the bottom of the tank at which the water jet will hit the ground is:
\[ x = 2 \sqrt{D(H - D)} \]