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 statioary on the ground are, respectively,

a, g

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(g - a); g

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a, a

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g, g

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Solution

The correct option is B $(g - a); g$
Acceleration of ball w.r.t man= acceleration of ball- acceleration of man

In first case, acceleration of man is a.
g>a
So, answer is (g-a)
In second case, acceleration of man is 0.
So answer is $g$

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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 statioary on the ground are, respectively,

The correct option is B (g−a);gAcceleration of ball w.r.t man= acceleration of ball- acceleration of manIn first case, acceleration of man is a. g>aSo, answer is (g-a)In second case, acceleration of man is 0. So answer is g

The correct option is B $(g - a); g$
Acceleration of ball w.r.t man= acceleration of ball- acceleration of man

In first case, acceleration of man is a.
g>a
So, answer is (g-a)
In second case, acceleration of man is 0.
So answer is $g$