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Step 1: Determine the Helicopter’s Velocity at Height h
Since the helicopter starts from rest and moves upward with constant acceleration $g$, its velocity when it reaches height $h$ can be found using the kinematic relation:
$ v^2 = u^2 + 2\,g\,h $
Here, $u = 0$ (starting from rest), so
$ v = \sqrt{2\,g\,h}.$
Step 2: Apply the Equation of Motion for the Dropped Packet
At the instant the packet is dropped, it has an initial upward velocity $v = \sqrt{2\,g\,h}$. Let downward be negative. The packet’s displacement from height $h$ to the ground is $-\,h$. Using
$ s = ut + \frac{1}{2}\,a\,t^2,$
we substitute $s = -\,h$, $u = \sqrt{2\,g\,h}$ (upward), and $a = -\,g$ (gravity acting downward). Thus,
$ -\,h = \sqrt{2\,g\,h}\,t - \frac{1}{2}\,g\,t^2.$
Step 3: Rearrange into a Quadratic Equation
Rearrange the above equation to:
$ g\,t^2 - 2\,\sqrt{2\,g\,h}\,t - 2\,h = 0.$
This is a quadratic equation in $t$.
Step 4: Solve the Quadratic Equation
Solving for $t$ yields an expression that simplifies to
$ t = 3.4\,\sqrt{\frac{h}{g}}.$
Step 5: Final Time to Reach the Ground
Hence, the time taken by the food packet to reach the ground is:
$ t = 3.4\,\sqrt{\frac{h}{g}}.$
