Helicopter Packet Problem – Detailed Solution

Helicopter Packet Problem

A helicopter rises from rest vertically upward with a constant acceleration \(2g\). A food packet is dropped from the helicopter when it is at a height \(h\) above the ground. Find the magnitude of the velocity with which the packet strikes the ground. (Here, \(g\) is the acceleration due to gravity.)

• (1) \(v = \sqrt{6gh}\)
• (2) \(v = \sqrt{3.4gh}\)
• (3) \(v = \sqrt{2gh}\)
• (4) \(v = \sqrt{3gh}\)

Select the Correct Option

(1) \(v = \sqrt{6gh}\)

(2) \(v = \sqrt{3.4gh}\)

(3) \(v = \sqrt{2gh}\)

(4) \(v = \sqrt{3gh}\)

Step‑by‑Step Explanation

Step 1: When the helicopter starts from rest and accelerates upward with acceleration \(2g\), its displacement \(h\) and velocity at the moment of drop are related by: \[ v_0^2 = u^2 + 2a\,h. \] Since \(u = 0\) and \(a = 2g\): \[ v_0^2 = 2(2g) h = 4gh \quad\Rightarrow\quad v_0 = 2\sqrt{gh}. \] This is the upward velocity of the helicopter—and hence the packet—when it is dropped.

Step 2: After being dropped, the packet is in free fall under gravity (with no additional acceleration besides \(g\) downward). Its initial velocity is upward \(v_0 = 2\sqrt{gh}\).

Step 3: Let the time taken for the packet to reach the ground be \(t\). Using the equation of motion (taking upward as positive): \[ y = v_0 t - \frac{1}{2} g t^2, \] where \(y = h\) (the displacement from drop height to ground, but here it will be \(-h\) since the packet falls downward relative to the drop point). Writing: \[ -h = 2\sqrt{gh}\,t - \frac{1}{2} g t^2. \] Rearranging: \[ \frac{1}{2} g t^2 - 2\sqrt{gh}\,t - h = 0. \]

Step 4: Solving this quadratic for \(t\) (using the positive time root) gives: \[ t = \frac{2\sqrt{gh} + \sqrt{(2\sqrt{gh})^2 + 2gh}}{g}. \] Notice that: \[ (2\sqrt{gh})^2 = 4gh \quad \text{and} \quad 4gh+2gh=6gh. \] Thus, \[ t = \frac{2\sqrt{gh} + \sqrt{6gh}}{g}. \]

Step 5: The velocity of the packet just before impact (taking downward as positive) is given by: \[ v = v_0 - g\,t. \] Substituting \(v_0 = 2\sqrt{gh}\) and \(t\) from above: \[ v = 2\sqrt{gh} - g\left(\frac{2\sqrt{gh} + \sqrt{6gh}}{g}\right) = 2\sqrt{gh} - \left(2\sqrt{gh} + \sqrt{6gh}\right) = -\sqrt{6gh}. \]

The negative sign indicates the final velocity is downward. The magnitude is: \[ v = \sqrt{6gh}. \]

Final Answer: The packet strikes the ground with speed \(v = \sqrt{6gh}\), which corresponds to option (1).

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