Relation Between Distances in Consecutive Time Intervals for Free Fall – Rankers Physics

Motion Under Gravity: Practice Problem & Solution

A stone falls freely under gravity. It covers distances \(h_1, h_2\) and \(h_3\) in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between \(h_1, h_2\) and \(h_3\) is: (2013)
\(h_1 = h_2 = h_3\)
\(h_1 = 2h_2 = 3h_3\)
\(h_1 = \frac{h_2}{3} = \frac{h_3}{5}\)
\(h_2 = 3h_1 and h_3 = 3h_2 \)

Solution Explained:

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

Concept: Distances covered by a freely falling body in equal successive time intervals.
Rule: For a body falling from rest, distances in successive equal time intervals are in ratio 1:3:5:...
Solution: \(h_1:h_2:h_3 = 1:3:5\). This implies \(h_2 = 3h_1\) and \(h_3 = 5h_1\). Therefore, \(h_1 = h_2/3 = h_3/5\).

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