Parachute Opening Height – Rankers Physics
Topic: Kinematics
Subtopic: Equations of Motion

Parachute Opening Height

A paratrooper jumps from a height \(H\). The parachute can provide a uniform deceleration of \(2\text{ ms}^{-2}\). The height above the ground at which the parachute should be opened so that he touches ground with zero speed is (take \(g = 10\text{ ms}^{-2}\)):
\(\frac{H}{6}\)
\(\frac{4H}{5}\)
\(\frac{5H}{6}\)
\(\frac{6H}{7}\)

Solution:

Let \(h\) be free fall and \(y\) be decelerating height, so \(H = h + y\). Speed before parachute opens: \(v^2 = 2gh = 20h\). Deceleration phase: \(0 = v^2 - 2ay = v^2 - 4y\), which gives \(20h = 4y \Rightarrow y = 5h\). Since \(H = 6h\), we find \(y = \frac{5H}{6}\).

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