Q. A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v? (Take g = 10 m/s²)
A) 60 m/s
B) 75 m/s
C) 55 m/s
D) 40 m/s
Explanation
[Solution from source] At t = 18 s the first ball has fallen ½ × 10 × 18² = 1620 m. The second ball moves for 12 s: 12v + ½ × 10 × 12² = 12v + 720. Equating gives 12v = 900, v = 75 m/s. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.