A balloon starts rising at an acceleration. After some time, a stone is dropped from it. If the stone reaches the ground in the same time in which balloon reached at this point from where stone is dropped, find the acceleration of the balloon:

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g/2

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g/3

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2g/3

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Solution

The correct option is C
g/3

$F o r b a l l o o n, v = a t S = \frac{1}{2} a t^{2} - S = v t^{'} - \frac{1}{2} g t^{' 2} F r o m e q n s . (1) a n d (2), - \frac{1}{2} a t^{2} = (a t) t^{'} - \frac{1}{2} g t^{' 2} G i v e n, + \frac{3}{2} a t^{2} = \frac{1}{2} g t^{2} a = \frac{g}{3}$

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A balloon starts rising at an acceleration. After some time, a stone is dropped from it. If the stone reaches the ground in the same time in which balloon reached at this point from where stone is dropped, find the acceleration of the balloon:

The correct option is C g/3 For balloon, v=atS=12at2−S=vt′−12gt′2From eqns.(1) and (2),−12at2=(at)t′−12gt′2Given, +32at2=12gt2a=g3

The correct option is C
g/3

$F o r b a l l o o n, v = a t S = \frac{1}{2} a t^{2} - S = v t^{'} - \frac{1}{2} g t^{' 2} F r o m e q n s . (1) a n d (2), - \frac{1}{2} a t^{2} = (a t) t^{'} - \frac{1}{2} g t^{' 2} G i v e n, + \frac{3}{2} a t^{2} = \frac{1}{2} g t^{2} a = \frac{g}{3}$