To solve the problem of finding the relationship between the ranges of the jets of water from the two holes in the tank, we can use principles from fluid mechanics, specifically Torricelli's law. This law states that the speed of efflux of a fluid under the force of gravity through an orifice is given by the equation: \( v = \sqrt{2gh} \), where \( v \) is the velocity of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the height of the fluid column above the hole.
Understanding the Setup
In this scenario, we have a water tank with a constant water level of depth \( 3a \). There are two holes in the tank:
- The first hole is located at a depth of \( a \) from the water level.
- The second hole is located at a depth of \( 2a \) from the water level.
Calculating the Heights
To find the heights \( h_1 \) and \( h_2 \) for the two holes, we can express them as follows:
- For the first hole: \( h_1 = 3a - a = 2a \)
- For the second hole: \( h_2 = 3a - 2a = a \)
Applying Torricelli's Law
Now, we can calculate the velocities of the water jets from each hole:
- Velocity from the first hole:
Using \( h_1 = 2a \), we have:
\( v_1 = \sqrt{2g(2a)} = \sqrt{4ga} = 2\sqrt{ga} \)
- Velocity from the second hole:
Using \( h_2 = a \), we find:
\( v_2 = \sqrt{2g(a)} = \sqrt{2ga} \)
Determining the Range of the Jets
The range \( R \) of a projectile launched horizontally from a height \( h \) can be calculated using the formula:
\( R = v \cdot t \), where \( t \) is the time of flight. The time of flight can be derived from the equation of motion under gravity:
\( h = \frac{1}{2}gt^2 \) implies \( t = \sqrt{\frac{2h}{g}} \).
For each hole, we can substitute the respective heights:
- For the first hole:
\( t_1 = \sqrt{\frac{2(2a)}{g}} = \sqrt{\frac{4a}{g}} = \frac{2\sqrt{a}}{\sqrt{g}} \)
Thus, the range \( R_1 = v_1 \cdot t_1 = (2\sqrt{ga}) \cdot \left(\frac{2\sqrt{a}}{\sqrt{g}}\right) = \frac{4a}{g} \)
- For the second hole:
\( t_2 = \sqrt{\frac{2a}{g}} \)
So, the range \( R_2 = v_2 \cdot t_2 = (\sqrt{2ga}) \cdot \left(\sqrt{\frac{2a}{g}}\right) = \frac{2a}{g} \)
Establishing the Relationship
Now we can compare the ranges:
- \( R_1 = \frac{4a}{g} \)
- \( R_2 = \frac{2a}{g} \)
From this, we can see that:
\( R_1 = 2R_2 \)
Final Thoughts
This means that the range of the jet from the first hole is twice that of the jet from the second hole. The difference in height of the holes directly affects the velocity of the water jets and, consequently, their ranges. Understanding these relationships helps in various applications, from designing water fountains to understanding fluid dynamics in engineering contexts.
