Q. A lift is moving down with acceleration$\ a$. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively
A) $g,g$
B) $g - a,\ g - a$
C) $g - a,\ g$
D) $a,\ g$
Explanation
(c) Apparent weight of ball
$w^{'} = w - R$
$R = ma\ $acts upward $= ma$
$w^{'} = mg - ma = m(g - a)$
Hence, apparent acceleration in the lift is $g - a$. Now if the man is standing stationary on the ground, then the apparent acceleration of the falling ball is $g.$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.