1. Velocity of the helicopter when the packet is dropped

The helicopter starts from rest and rises upward with constant acceleration $g$.

Using $v^2=u^2+2as$ with $u=0$, $a=g$, and displacement $s=h$, we get $v^2=2gh$ so the upward velocity of the packet at the instant of release is $v=\sqrt{2gh}.$

2. Motion of the packet after release

After being dropped, the packet has:

• initial height $h$
• initial upward velocity $\sqrt{2gh}$
• downward acceleration $g$

Take upward as positive. Then the position of the packet is $y=h+vt-\frac{1}{2}gt^2$ When it reaches the ground, $y=0$. Therefore, $0=h+\sqrt{2gh}\,t-\frac{1}{2}gt^2$

3. Solve the quadratic equation

Rearranging, $\frac{1}{2}gt^2-\sqrt{2gh}\,t-h=0$ Multiply by $2$: $gt^2-2\sqrt{2gh}\,t-2h=0$

Using the quadratic formula, $t=\frac{2\sqrt{2gh}\pm\sqrt{(2\sqrt{2gh})^2+8gh}}{2g}$ Now, $(2\sqrt{2gh})^2=8gh$ so $t=\frac{2\sqrt{2gh}\pm\sqrt{8gh+8gh}}{2g}$ $t=\frac{2\sqrt{2gh}\pm\sqrt{16gh}}{2g}$ $t=\frac{2\sqrt{2gh}\pm4\sqrt{gh}}{2g}$

We take the positive root: $t=\frac{2\sqrt{2gh}+4\sqrt{gh}}{2g}$ $t=\frac{\sqrt{gh}(2\sqrt2+4)}{2g}$ $t=(\sqrt2+2)\sqrt{\frac{h}{g}}$

4. Numerical value

Since $\sqrt2+2\approx1.414+2=3.414$ we get $t\approx3.4\sqrt{\frac{h}{g}}.$

5. Match with options

This matches Option A.

Derived answer: $\boxed{A}$