A lift is moving down with an acceleration a. | vedclass

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(C) $1$. For the man standing stationary on the ground (inertial frame),the ball is under the influence of gravity alone. Therefore,its acceleration i

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$g - a, g$

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(C) $1$. For the man standing stationary on the ground (inertial frame),the ball is under the influence of gravity alone. Therefore,its acceleration is $g$ downwards.$2$. For the man inside the lift (non-inertial frame),we must apply a pseudo force. The lift is accelerating downwards with $a$,so a pseudo force $ma$ acts upwards on the ball.$3$. The net force on the ball in the lift frame is $F_{net} = mg - ma$ (downwards).$4$. The acceleration observed by the man in the lift is $a_{lift} = F_{net} / m = (mg - ma) / m = g - a$ downwards.$5$. Thus,the accelerations are $g - a$ and $g$ respectively.

$A$ lift is moving down with an 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:

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