91Ó°ÊÓ

Kinematics is the study of motion without considering the forces that cause it. In our projectile motion problem involving the airplane, key kinematic principles apply. We are interested in how objects like the package move through the air. This often requires us to predict both the position and velocity at various points in its trajectory. The movement of the package can be split into two independent components: horizontal and vertical motion. This separation simplifies calculations because while the vertical motion is affected by gravity, the horizontal motion is not. Understanding this allows us to analyze the problem effectively. Remember that kinematics helps us to predict where and when an object will be at any given point during its flight.

Trigonometry is essential in physics, especially when dealing with angles and directions. In this problem, the aircraft is climbing at an angle of \(50.0^\circ\). This angle affects how we determine the velocity components. Trigonometric functions such as cosine and sine are used to resolve the airplane's velocity into horizontal and vertical components.

• The horizontal component \(v_x\) reflects the velocity parallel to the ground.
• The vertical component \(v_y\) represents the velocity perpendicular to the ground.

These components are crucial for solving projectile motion problems. By using \( \cos(\theta) \) for horizontal and \( \sin(\theta) \) for vertical, where \( \theta \) is the angle with the horizontal, we can determine how much the velocity affects each direction.

Breaking down velocity into components makes complex motion easier to analyze. For the airplane, its speed \(v = 97.5 \text{ m/s}\) is divided into horizontal and vertical components using trigonometric functions:

• \(v_x = v \cdot \cos(50.0^\circ)\)
• \(v_y = v \cdot \sin(50.0^\circ)\)

Once these components are calculated, they give a clearer picture of how the package will move both horizontally and vertically after release. The horizontal velocity component \(v_x\) helps to determine how far along the ground the package travels, while the vertical component \(v_y\) affects how long it stays in the air. Understanding these is vital for predicting distance and time of flight in projectile motion.

The quadratic formula is a powerful tool for solving equations involving squared terms. In our exercise, it helps to determine the time of flight, \(t\), for the package once it's released. The vertical motion equation \(y = v_y \cdot t - \frac{1}{2} g t^2\) describes how the package's altitude changes over time. By setting \(y = -732 \text{ m}\) and substituting known values into the equation, we get a quadratic in the form \(at^2 + bt + c = 0\).
The quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) is then used to solve for \(t\). This time \(t\) reveals how long the package remains airborne, allowing for further calculation of its horizontal distance and velocity angle upon impact.