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Question

A body is launched from Earth at 12.5km/s. What is its speed at infinity? (Escape speed = 11.2km/s)

Options

Choose one · Correct answer highlighted

Explanation

vf2 = vi2−ve2. vf2 = (12.5)2−(11.2)2 = 156.25−125.44 = 30.81. vf = 30.81≈5.55km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 5.5 km/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

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