Question:
A helicopter reises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is a height $h$. The time taken by the packet to reach the ground is close to $[g$ is the acceleration due to gravity] :
Correct Option: , 3
Solution:
$v^{2}=u^{2}+2 a s$
$v^{2}=2 g h+2 g h$
$v=\sqrt{4 g h}$
$\Rightarrow \sqrt{4 \mathrm{gh}}=\sqrt{2 \mathrm{gh}}+\mathrm{gt}$
$\Rightarrow \mathrm{t}=\sqrt{\frac{4 \mathrm{~h}}{\mathrm{~g}}}-\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \Rightarrow 3.4 \sqrt{\frac{\mathrm{h}}{\mathrm{g}}}$