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

Pranav
May 26, 2024, 6:35 a.m.
If the helicopter leaves the packet then the packet will also move upwards for some time then it will start comming down Then how are you calculating just the time from the position where it was dropped to ground
Rohit
May 17, 2024, 6:35 a.m.
Hi Sarah Sir ðŸ‘‹
sarah
May 21, 2024, 6:35 a.m.
hi beta
April 5, 2023, 6:35 a.m.
How h is negative
Ayesha
Sept. 30, 2023, 6:35 a.m.
Becoz we have taken downward direction as - ve
Manasvi Bhushan
Sept. 14, 2022, 9:29 a.m.
sir koi dus ra approach baato na ye to sab soch lete hai