A helicopter rises from rest on the ground vertically upwards with a constant acceleration g.

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 accelertion due to gravity] :

  1. (1) $t=\frac{2}{3} \sqrt{\left(\frac{h}{g}\right)}$

  2. (2) $t=1.8 \sqrt{\frac{h}{g}}$

  3. (3) $t=3.4 \sqrt{\left(\frac{h}{g}\right)}$

  4. (4) $t=\sqrt{\frac{2 h}{3 g}}$


Correct Option: , 3

Solution:

(3) For upward motion of helicopter,

$v^{2}=u^{2}+2 g h \Rightarrow v^{2}=0+2 g h \Rightarrow v=\sqrt{2 g h}$

Now, packet will start moving under gravity.

Let ' $t$ ' be the time taken by the food packet to reach the ground.

$s=u t+\frac{1}{2} a t^{2}$

$\Rightarrow-h=\sqrt{2 g h} t-\frac{1}{2} g t^{2} \Rightarrow \frac{1}{2} g t^{2}-\sqrt{2 g h} t-h=0$

or, $t=\frac{\sqrt{2 g h} \pm \sqrt{2 g h+4 \times \frac{g}{2} \times h}}{2 \times \frac{g}{2}}$

or, $t=\sqrt{\frac{2 g h}{g}}(1+\sqrt{2}) \Rightarrow t=\sqrt{\frac{2 h}{g}}(1+\sqrt{2})$

or, $t=3.4 \sqrt{\frac{h}{g}}$

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Manasvi Bhushan
Sept. 14, 2022, 9:29 a.m.
sir koi dus ra approach baato na ye to sab soch lete hai