Question

A lift is moving down with acceleration a , A man un 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 . [AIEEE 2002] ​

Answer 1

Answer:

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Explanation:

Apparent weight of the ball is W'= W-R

R = ma (Acts upward)

W' = mg - ma = m(g-a)

Hence, the apparent acceleration in the lift is g-a and for the person standing outside the acceleration of the falling ball is g.

Answer 2

Explanation:- Apparent weight of ball

• W' = w-R [Where, R = normal reaction]
• R = ma (acting upward]
• w' = mg - ma = m(g-a)

Hence, apprent 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 .