Question

An airplane travels horizontally at a constant velocity v⃗ .An object is dropped from the plane, and one second later another object is dropped from the plane. If air resistance is negligible, what happens to the vertical distance between the two objects while they are both falling?

Answer 1

time increases, the distance between the two bodies becomes less until, for long periods of time, the two bodies are almost at the same height.

Step-by-step explanation:

Give us the solution to this exercise, using kinematics

Let's use index 1 for the body that falls first and index 2 for the body that leaves 1 second later. Let's use the same timer

y₁ = y₀ + go t - ½ g t²

y₂ = y₀ + go t - ½ g (t-1)²

As the two bodies come to rest their initial vertical velocity is zero.

y₁-y₀ = - ½ g t²

y₂ -y₀ = - ½ g (t-1)²

y₀ - y₁ = 4.9 t²

y₀-y₂ = 4.9 (t-1)²

The relationship between the two distance is

y₀-y₂ / y₀-y₁ = [(t-1) / t]²

y₀-y₂ / y₀-y₁ = [1-1 / t]²

Therefore, as time increases, the distance between the two bodies becomes less until, for long periods of time, the two bodies are almost at the same height.

Answer 2

Step-by-step explanation:

Assuming G as gravitational force. Free fall has been demonstrated to depend only by the gravity force. Since both objects are thrown from the same plane, and assuming wind resistance 0. The distance between objects is the same since gravity is the same.