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Understanding the concept of net force is crucial when analyzing rocket dynamics. Net force is the overall force acting on an object when taking into account all the individual forces applied to it. In essence, it's the sum of all the vectors of the forces, which determines how an object will move.

For a rocket, two main forces are at play: thrust, which is the force propelling the rocket upwards, and gravity, which pulls it downward. The net force (\( F_{net} \)) is therefore calculated by subtracting the gravitational force from the thrust. For the given model rocket:

• Thrust (\( F_{thrust} \)): 25 N
• Weight/Gravitational force (\( F_{gravity} \)): 13.72 N

So,\[ F_{net} = F_{thrust} - F_{gravity} = 25 \, \text{N} - 13.72 \, \text{N} = 11.28 \, \text{N} \]This net force is what determines how the rocket accelerates upwards through Newton’s Second Law of Motion.

Acceleration is a vital element in understanding how quickly a rocket changes its velocity over time. According to Newton's Second Law of Motion, the acceleration (\( a \)) of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

The relationship can be described by the formula:\[ F_{net} = m \cdot a \]Rearranging the formula to solve for acceleration gives:\[ a = \frac{F_{net}}{m} \]For the model rocket, with a net force of 11.28 N and mass of 1.4 kg:\[ a = \frac{11.28}{1.4} = 8.06 \, \text{m/s}^2 \]This indicates the rocket's rate of velocity increase due to the net force acting on it while under thrust. Understanding how to calculate this helps predict the velocity and position of the rocket at different points in time.

Equations of motion are employed to understand and predict the future state of a moving object, given its initial conditions. They help calculate various parameters such as displacement, velocity, and time. For the model rocket scenario, several equations of motion are used:

1. **Calculating the final velocity when thrust ends:**
Using: \[ v^2 = u^2 + 2as \] Here, \( u = 0 \, \text{m/s} \) (the initial velocity), \( a = 8.06 \, \text{m/s}^2 \), and \( s = 15 \, \text{m} \). Solving gives: \[ v = \sqrt{0 + 2 \times 8.06 \times 15} = 15.55 \, \text{m/s} \].
2. **Calculating maximum height reached after thrust ended:**
Once the thrust ends, the rocket continues to ascend due to its velocity until gravity decelerates it to zero: \[ 0 = (15.55)^2 + 2(-9.8)s \] Simplifying gives: \[ s = \frac{(15.55)^2}{2 \times 9.8} = 12.35 \, \text{m} \].
The total maximum height is 27.35 m (15 + 12.35 m).

3. **Calculating the speed upon returning to the ground:**
Given the absence of air resistance, the rocket's speed returning will equal its speed at the end of the thrust phase: \[ v = 15.55 \, \text{m/s} \].
These equations not only help determine key aspects of the rocket's journey but also demonstrate the interconnectedness of physics principles in motion analysis.