Question 58 of 308

A 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. The time taken by the packet to reach the ground is close to [$g$ is the acceleration due to gravity]:

A$t=\dfrac{2}{3}\sqrt{\left(\dfrac{h}{g}\right)}$B$t=1.8\sqrt{\left(\dfrac{h}{g}\right)}$C$t=3.4\sqrt{\left(\dfrac{h}{g}\right)}$D$t=\sqrt{\dfrac{2h}{3g}}$

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Answer: (C) $t=3.4\sqrt{\left(\dfrac{h}{g}\right)}$

$\Rightarrow$For upward motion of helicopter
${v}^{2}={u}^{2}+2as$
${v}^{2}=0+2gh$
$v=\sqrt{2gh}$
$\Rightarrow$ Now particle will start moving under gravity.
$s=ut+\dfrac{1}{2}a{t}^{2}$
$-h=\sqrt{2gh}t-\dfrac{1}{2}g{t}^{2}$
then $t=\dfrac{\sqrt{2gh}\pm \sqrt{2gh+4\times \dfrac{g}{2}\times h}}{2\times \dfrac{g}{2}}$
$t=\sqrt{\dfrac{2gh}{g}}(1+\sqrt{2})$
$t=\sqrt{\dfrac{2h}{g}}(1+\sqrt{2})$