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Laws of Motion Part 1

Section: Physics

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Question : 95 of 100

Marks: +1, -0

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

$g , g$
$g-a, g-a$
$g-a, g$
$a, g$

Solution:

Let acceleration of ball

=\vec{a}_b

and acceleration of man is

=\vec{a}_m

With respect to the man standing in the lift, the acceleration of the ball

\vec{a}_{bm} = \vec{a}_b - \vec{a}_m

\Rightarrow a_{bm} = g - a

Where

a

is the acceleration of the man as the acceleration of the lift is

a

.

With respect to the man standing on the ground the acceleration of the ball

\vec{a}_{bm} = \vec{a}_b - \vec{a}_m

\Rightarrow a_{bm} = g - 0

=g

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