Question

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

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

Answer 1

Answer: (C)

$v^{2}=u^{2}+2 a s$
$v^{2}=2 g h+2 g h$
$v=\sqrt{4 gh}$
$\Rightarrow \sqrt{4 gh }=\sqrt{2 gh }+ gt$
$\Rightarrow t =\sqrt{\frac{4 h }{ g }}-\sqrt{\frac{2 h }{ g }}
\Rightarrow 3.4 \sqrt{\frac{ h }{ g }}$