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

Question

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

Tardigrade Admin · Accepted Answer

Apparent weight of ball w prime=w-R R=m a (acts upward) w prime=m g-m a=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.

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

$g, g$

$g - a, g - a$

$g - a, g$

$a, g$

Apparent weight of ball
$w^{'} = w - R$
$R = ma$ (acts upward)
$w^{'} = m g - ma = m (g - a)$
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 $g$ .

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

g,g

g−a,g−a

g−a,g

a,g