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 $t = 18s$. What is the value of $v$? (taking $g = 10\ m/s^{2}$)
A) $60\ m/s$
B) $75\ m/s$
C) $55\ m/s$
D) $40\ m/s$
Explanation
(b) Let two balls meet at depth $h$ from platform
So $h = \frac{1}{2}g(18)^{2} = v(12) + \frac{1}{2}g(12)^{2} \Rightarrow v = 75\ ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.