Helicopters to the rescue!

The less-than-obvious element to this problem is that at the point that the food packet is dropped, it has an upwards velocity thanks to the helicopter. We begin by calculating the velocity of the helicopter at the point at which the food packet is dropped.

We are told that the helicopter has a constant upwards acceleration of $\frac{g}{8}$ and has initial displacement and velocity of $0$ . This means that its height above the ground is: $\begin{aligned} d(t) &= \iint \frac{g}{8} \mathrm{dt dt} \\ &= d(0) + v(0)t + \frac{gt^2}{16} \\ &= \frac{gt^2}{16} \end{aligned}$ The time at which the packet is dropped is $t_h$ , which is when $d(t_h) = h$ . Solving gives: $\begin{aligned} \frac{gt_h^2}{16} &= h \\ t_h^2 &= \frac{16h}{g} \\ t_h &= 4\sqrt{\frac{h}{g}} \end{aligned}$ The velocity of the helicopter is the derivative of its displacement. That is, $v(t) = \frac{gt}{8}$ . The initial velocity of the food packet is the same as the velocity of the helicopter at $t_h$ . That is, $\begin{aligned} v(t_h) &= \frac{gt_h}{8} \\ &= 4\sqrt{\frac{h}{g}}\cdot\frac{g}{8} \\ &= \frac{\sqrt{gh}}{2} \end{aligned}$ We now set up the equation of motion for the food packet. It's initial displacement is $h$ , and its initial velocity is $v(t_h)$ as above. It experiences constant acceleration of $-g$ , so its equation for the displacement of the food packet at time $t$ , $d_p(t)$ , is: $d_p(t) = h + \frac{\sqrt{gh}}{2}t - \frac{g}{2}t^2$ Note that this equation is for $t$ seconds after the food packet is dropped, which is different from previous equations. As a consequence, the time taken for the food packet to reach the ground is the time $t_p$ such that $d_p(t_p) = 0$ . That is: $\begin{aligned} t_p &= \frac{ -\frac{\sqrt{gh}}{2} \pm \sqrt{ \frac{gh}{4} + 2gh}}{-g} \\ &= \frac{1}{2}\sqrt{\frac{h}{g}} \mp \sqrt{\left(\frac{1}{4} + 2\right)\frac{h}{g}} \\ &= \frac{1}{2}\sqrt{\frac{h}{g}} \mp \sqrt{\frac{9}{4}\cdot\frac{h}{g}} \\ &= \left(\frac{1}{2} \mp \frac{3}{2}\right)\sqrt{\frac{h}{g}} \\ &= \frac{1 \mp 3}{2}\sqrt{\frac{h}{g}} \end{aligned}$ Since we are clearly only interested in the positive solution, we must have: $t_p = \boxed{2\sqrt{\frac{h}{g}}}$