If a helicopter is moving upwards with acceleration `g` and food packet are thrown from a height `h`. Find time period to reach those packet to ground

Answer

To solve the problem of finding the time it takes for food packets thrown from a helicopter moving upwards with acceleration \( g \) to reach the ground from a height \( h \), we can follow these steps:

### Step 1: Determine the velocity of the helicopter at height \( h \)

The helicopter is moving upwards with an acceleration \( g \). We can use the kinematic equation:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) is the final velocity,
- \( u \) is the initial velocity (0, since it starts from rest),
- \( a \) is the acceleration (\( g \)),
- \( s \) is the displacement (\( h \)).

Substituting the values:

\[ v^2 = 0 + 2gh \]

Thus, the final velocity \( v \) when the helicopter reaches height \( h \) is:

\[ v = \sqrt{2gh} \]

### Step 2: Analyze the motion of the food packet after it is released

When the food packet is released from the helicopter, it has an initial upward velocity \( u = \sqrt{2gh} \) and will be acted upon by gravity, which will decelerate it.

### Step 3: Set up the equation of motion for the food packet

The food packet will first move upwards until its velocity becomes zero and then it will start falling down. We need to find the time it takes for the packet to reach the ground.

Using the kinematic equation for displacement:

\[ s = ut + \frac{1}{2} a t^2 \]

Where:
- \( s \) is the total displacement (which will be \( -h \) since it moves from height \( h \) to the ground),
- \( u = \sqrt{2gh} \) (initial velocity),
- \( a = -g \) (acceleration due to gravity acting downwards),
- \( t \) is the time taken.

Substituting the values:

\[ -h = \sqrt{2gh} \cdot t - \frac{1}{2} g t^2 \]

### Step 4: Rearranging the equation

Rearranging gives us:

\[ \frac{1}{2} g t^2 - \sqrt{2gh} \cdot t - h = 0 \]

This is a quadratic equation in \( t \) of the form \( at^2 + bt + c = 0 \), where:
- \( a = \frac{1}{2} g \),
- \( b = -\sqrt{2gh} \),
- \( c = -h \).

### Step 5: Solve the quadratic equation

Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ t = \frac{-(-\sqrt{2gh}) \pm \sqrt{(-\sqrt{2gh})^2 - 4 \cdot \frac{1}{2} g \cdot (-h)}}{2 \cdot \frac{1}{2} g} \]

This simplifies to:

\[ t = \frac{\sqrt{2gh} \pm \sqrt{2gh + 2gh}}{g} \]

\[ t = \frac{\sqrt{2gh} \pm \sqrt{4gh}}{g} \]

\[ t = \frac{\sqrt{2gh} \pm 2\sqrt{gh}}{g} \]

### Step 6: Choose the appropriate solution

Since time cannot be negative, we take the positive root:

\[ t = \frac{\sqrt{2gh} + 2\sqrt{gh}}{g} \]

Factoring out \( \sqrt{gh} \):

\[ t = \frac{\sqrt{gh}(\sqrt{2} + 2)}{g} \]

### Final Result

Thus, the time taken for the food packets to reach the ground is:

\[ t = \frac{(\sqrt{2} + 2) \sqrt{h}}{\sqrt{g}} \]