Practice question
Question
A body is launched from Earth at 14km/s. What is its speed at infinity? (Escape speed = 11.2km/s)
Explanation
vf2 = vi2−ve2. vf2 = (14)2−(11.2)2 = 196−125.44 = 70.56. vf = 70.56≈8.4km/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 8.4 km/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Discussion
Comments
Share your thoughts. New comments appear after admin approval.
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.