A body is projected from Earth with 13km/s. What is its speed far away? (Escape speed = 11.2km/s)
12vi2−12ve2 = 12vf2. vf2 = (13)2−(11.2)2 = 169−125.44 = 43.56. vf = 43.56≈6.6km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.6 km/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.
Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.