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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]
$t =\sqrt{\frac{2 h }{3 g }}$
$t=1.8 \sqrt{\frac{h}{g}}$
$t=3.4 \sqrt{\left(\frac{h}{g}\right)}$
$t=\frac{2}{3} \sqrt{\left(\frac{h}{g}\right)}$
Solution
$t_{0}=\sqrt{\frac{2 h}{g}}$
$\Rightarrow \quad u=\sqrt{\frac{2 h}{g}} \times g=\sqrt{2 g h}$
$\therefore \quad t_{1}=$time to reach top $=\frac{u}{g}=\sqrt{\frac{2 h}{g}}$
$\therefore \quad H=h+h^{\prime}=2 h$
$t_{2}=$ time of fall $=\sqrt{\frac{2 \times(2 h)}{g}}$
$\quad=2 \sqrt{\frac{h}{g}}$
Total time $=t_{1}+t_{2}$
$\quad=(2+\sqrt{2}) \sqrt{\frac{h}{g}}$
$\quad=3.4 \sqrt{\frac{h}{g}}$
