Packet from a plane. A plan flies at constant velocity and altitude, and drops a packet. Describe trajectory of the packet as seen from the ground and from the plane.

When we talk about projectile motion, we are discussing objects that are thrown or propelled into the air and are subject to gravity and air resistance. In this scenario, the packet dropped from the plane is a classic example of projectile motion. The packet is subject to two forces: the constant horizontal velocity of the plane, and the pull of gravity, which accelerates it downward. This combination of forces results in a unique trajectory.

It's crucial to understand that the horizontal motion and the vertical motion of the packet are independent of each other. The packet will keep moving forward at the same speed as the plane while simultaneously being pulled downward by gravity.

• Horizontal motion is uniform (constant velocity).
• Vertical motion is uniformly accelerated (due to gravity).

These independent motions combine to create the parabolic trajectory observed from the ground.

From the ground, the packet's path looks parabolic. This is because we observe the interplay between the packet’s horizontal motion (maintained since it was dropped from a moving plane) and its vertical acceleration due to gravity.

Mathematically, the trajectory of the packet dropped from the plane is a parabola, and it can be described by the equations:

• Horizontal position: \( x(t) = v_{\text{plane}} \times t \)
• Vertical position: \( y(t) = h - \frac{1}{2} g t^2 \)

Where:

• \(v_{\text{plane}}\) is the plane's velocity
• \(h\) is the altitude from which the packet is dropped
• \(g\) is the acceleration due to gravity
• \(t\) is the time elapsed

This vertical motion showcases how gravity pulls the packet down over time, while the constant horizontal velocity keeps the packet moving forward, resulting in the downward curved path.

Observing motion from different reference points helps us understand relative motion. In this exercise, we look at the packet from the perspectives of an observer on the ground and someone inside the plane.

From the ground, the packet’s path forms a parabola because we can see both the horizontal motion (matching the plane's speed) and the vertical acceleration (gravity pulling the packet down).

From the plane, the packet seems to just fall straight down. This is because an observer inside the plane moves with the same horizontal speed. Thus, they only see the vertical motion of the packet, giving the illusion that it’s falling straight down without moving forward. This difference in observation highlights the concept of relative motion, where the observed trajectory changes depending on the reference point.

Understanding the displacements in horizontal and vertical directions is key to grasping the overall motion of the packet.

**Horizontal Displacement**
The horizontal displacement of the packet depends on the constant velocity of the plane. Since there's no acceleration in the horizontal direction (the velocity is constant), we can simply express the horizontal position over time as:

• \( x(t) = v_{\text{plane}} \times t\)

**Vertical Displacement**
In contrast, the vertical displacement involves acceleration due to gravity. As the packet falls, it accelerates downward, and its vertical position over time can be expressed as:

• \( y(t) = h - \frac{1}{2} g t^2\)

Here, the packet starts at height \( h \) and its downward movement speed increases over time due to gravity (\( g \)). The combination of these displacements results in the parabolic path we observe from the ground.