Question

A helicopter is rising up from ground with an acceleration of $g{\text{m/s}}^2$, starting from rest after raising a height $h$ it attains a velocity of $v\text{m/s}$. At this instant, a particle is now released from the helicopter. Take $t=0$ at releasing time, calculate the time $t$ when particle reaches to the ground.

• A. $\sqrt{\frac{2h}{g}}$
• B. $2\sqrt{\frac{2h}{g}}$
• C. $(1+\sqrt{2})\sqrt{\frac{2h}{g}}$
• D. $4\sqrt{\frac{2h}{g}}$

Answer 1

The correct option is C $(1+\sqrt{2})\sqrt{\frac{2h}{g}}$

For upward motion of helicopter:

From third equation of motion,

$v^2=u^2+2as$

Here, $u=0;v=v;a=g;s=h$

$v^2=0+2as$

$v=\sqrt{2gh}$

Now, for the particle, motion will be as shown in figure, and it starts moving under gravity.


So, from second equation of motion,

$s_1=u_{1}t+\frac{1}{2}{a}_1{t}^2$

Here, $u_1=v=\sqrt{2gh};s_1=-h;a_1=-g$

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

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

So, value of $t$ will be,

$t=\frac{\sqrt{2gh}\pm \sqrt{2gh-(4\times \frac{g}{2}\times (-h)})}{2\times \frac{g}{2}}$

$t=\sqrt{\frac{2h}{g}}(1\pm \sqrt{2})$

Since, $t$ can't be negative,

$\therefore t=\sqrt{\frac{2h}{g}}(1+\sqrt{2})$

Hence, option $\left(\text{C}\right)$ is correct.