Let the ground be an inertial frame of reference S, and the airplane (pilot) be an inertial frame of reference S'. The airplane moves with a constant horizontal velocity $\vec{v}_p = v_x \hat{i}$ relative to the ground. The vertical direction is represented by the y-axis.

At the moment of release ($t=0$), the object is at the same position as the pilot. Let's set this release point as the origin $(0,0)$ in the pilot's frame, and at an altitude $H$ in the ground frame. Due to inertia, the object's initial velocity is identical to that of the airplane: $\vec{v}_{o, \text{initial}} = v_x \hat{i}$.

[Q1] From the pilot's perspective, what kind of motion does the dropped object undergo?

1. Motion in the Ground Frame (S) The motion of the pilot is uniform:

$$x_p(t) = v_x t$$ $$y_p(t) = H$$

The motion of the dropped object is projectile motion, as it is only subject to gravitational acceleration $\vec{a} = -g \hat{j}$:

$$x_o(t) = v_x t$$ $$y_o(t) = H - \frac{1}{2}gt^2$$

2. Motion in the Pilot's Frame (S') The motion of the object relative to the pilot is found by taking the difference between their coordinates in the ground frame. Let the relative coordinates be $(\Delta x, \Delta y)$.

Relative horizontal position:

$$\Delta x(t) = x_o(t) - x_p(t)$$ $$\Delta x(t) = v_x t - v_x t = 0$$

Relative vertical position:

$$\Delta y(t) = y_o(t) - y_p(t)$$ $$\Delta y(t) = \left(H - \frac{1}{2}gt^2\right) - H = -\frac{1}{2}gt^2$$

3. Conclusion The relative position vector of the object with respect to the pilot is $\vec{r}_{\text{rel}}(t) = (0) \hat{i} - (\frac{1}{2}gt^2) \hat{j}$.

Since the relative horizontal position $\Delta x$ is always zero, the object remains directly below the pilot at all times. The relative vertical position $\Delta y = -\frac{1}{2}gt^2$ describes an object starting from rest and accelerating downwards at a constant rate $g$.

Therefore, from the pilot's perspective, the object undergoes vertical free-fall motion.