The helicopter rises from rest on the ground vertically upwards with a constant acceleration g . A food packet is dropped from the helicopter when it is at a height h. Time taken by the packet to reach the ground is closed to [ g is the acceleration due to gravity]

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$t=3.4 \sqrt{(\frac{h}{g})}$

$t= \sqrt{\frac{2h}{3g}}$

$t=1.8 \sqrt{\frac{h}{g}}$

$t= \frac{2}{3} \sqrt{(\frac{h}{g})}$

answer is A.

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Detailed Solution

$\begin{array}{l} P a c k e t w i l l h a v e s a m e v e l o c i t y a s t h a t o f h e l i c o p t e r w h e n i t i s d r o p p e d . \\ U s e E q u a t i o n o f K i n e m a t i c s b e t w e e n p o i n t o f d r o p a n d g r o u n d . \\ -h= \sqrt{2gh} t- \frac{1}{2} {gt}^{2} \\ On Solving t= (\sqrt{2} +2) \sqrt{\frac{h}{g}} =3.4 \sqrt{\frac{h}{g}} \end{array}$