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

Answer 1

The correct option is C $t=3.4\sqrt{\left(\frac{h}{g}\right)}$

For upward motion of helicopter,
$v^2=u^2+2gh\Rightarrow v^2=0+2gh\Rightarrow v=\sqrt{2gh}$
Now, packet will start moving under gravity.

Let ${}^{\prime }{t}^{\prime }$ be the time taken by the food packet to reach the ground.
$s=ut+\frac{1}{2}a{t}^2$
$\Rightarrow -h=\sqrt{2gh}t-\frac{1}{2}g{t}^2\Rightarrow \frac{1}{2}g{t}^2-\sqrt{2gh}t-h=0$
$t=\frac{\sqrt{2gh}\pm \sqrt{2gh+4\times \frac{g}{2}\times h}}{2\times \frac{g}{2}}$
$t=\sqrt{\frac{2gh}{g}}(1+\sqrt{2})\Rightarrow t=\sqrt{\frac{2h}{g}}(1+\sqrt{2})$
$t=3.4\sqrt{\frac{h}{g}}$