Relative Distance in Free Fall – Rankers Physics

Relative Motion in One Dimension: Practice Problem & Solution

Two bodies are held separated by \(9.8\text{ m}\) vertically one above the other. They are released simultaneously to fall freely under gravity. After \(2\text{ s}\) the relative distance between them is :
4.9 m
19.6 m
9.8 m
39.2m

Solution Explained:

To solve this problem, we apply the core principles of Relative Motion in One Dimension. Understanding the underlying formula is key to arriving at the correct answer below:

Both bodies are released simultaneously and fall under gravity. Their acceleration is identical (\(g\)). Since their initial relative velocity is zero and relative acceleration is zero, their relative distance remains constant. Thus, after \(2\) s, the relative distance is still \(9.8\) m.

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