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When dealing with problems involving projectile motion, one of the most essential steps is decomposing the initial velocity of the object. The initial velocity in a projectile problem is typically specified by its magnitude and the angle of projection from the horizontal.

In the given problem, the initial velocity is 300 m/s at an angle of 55.0° above the horizontal. To solve for the motion of the projectile, we first decompose this initial velocity into two perpendicular components:

• Horizontal component \((v_{0x})\): This component acts along the horizontal direction. It is found using the formula \(v_{0x} = v_0 \cdot \cos(\theta)\), where \(v_0\) is the initial velocity and \(\theta\) is the angle above the horizontal.
  
• Vertical component \((v_{0y})\): This component acts along the vertical direction. It can be determined using \(v_{0y} = v_0 \cdot \sin(\theta)\).

By decomposing the initial velocity into these components, we can analyze and calculate various aspects of the projectile's motion along each direction independently.

Once the initial velocity components have been determined, they allow for the calculation of the projectile's motion in the horizontal and vertical directions separately. Understanding and using these components is key to solving projectile motion problems.

• Horizontal Motion: Since there is no air resistance or any other horizontal force acting on the projectile, the horizontal velocity remains constant. Thus, the horizontal distance, or x-coordinate, can be calculated using the formula \(x = v_{0x} \cdot t\). In this equation, \(v_{0x}\) is the constant horizontal velocity, and \(t\) is the time of flight.
  
• Vertical Motion: In contrast, the vertical motion is influenced by gravity. This results in a changing vertical velocity over time. The vertical position, or y-coordinate, of the projectile can be found with the equation \(y = v_{0y} \cdot t - 0.5 \cdot g \cdot t^2\), where \(g\) is the acceleration due to gravity, equal to 9.80 m/s².
  

By understanding these components and how they operate, you can accurately predict the trajectory, range, and height of the projectile.

Trajectory analysis involves understanding how a projectile moves through its path, defined by the curved trajectory. By analyzing the path, we can determine significant points such as the object's maximum height and the point where it lands.

In this specific problem, since the projectile explodes after 42.0 seconds, we can use the trajectory equations derived from the horizontal and vertical components to find the specific \(x\) and \(y\) coordinates at this time. This requires substituting the initial velocity components and time of flight into the corresponding equations:

• The horizontal distance is found by \(x = v_{0x} \cdot 42\).
• The vertical distance is found by \(y = v_{0y} \cdot 42 - 0.5 \cdot 9.80 \cdot 42^2\).

These calculations provide insight into where along its path the projectile will be found at any point in time, particularly when it reaches the 42-second mark. Understanding trajectory is crucial in applications across physics, engineering, and even in real-world scenarios such as sports and military operations.