A helicopter is flying horizontally with a speed 'v' at an altitude V has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped ?

Options

A. `sqrt(2v^(2)h)/(g )+h^(2))`
B. `(sqrt(2ghv^(2)+1)/(h^(2))`
C. `sqrt((2hv^(2))/(g)+h^(2))`
D. `sqrt(2gh)/(v^(2))+h^(2)`

Answer

To solve the problem of determining the distance of the helicopter from the man on the ground when the food packet is dropped, we can follow these steps:

### Step 1: Understand the scenario
The helicopter is flying horizontally at an altitude \( H \) with a horizontal speed \( V \). When the food packet is dropped, it will take some time to reach the ground due to gravity.

### Step 2: Calculate the time taken for the packet to fall
The vertical motion of the food packet can be described by the equation of motion under gravity. The initial vertical speed of the packet is \( 0 \) (since it is dropped), and it falls a distance \( H \). The equation for the distance fallen under gravity is:
\[ H = \frac{1}{2} g t^2 \]
From this, we can solve for \( t \):
\[ t = \sqrt{\frac{2H}{g}} \]

### Step 3: Calculate the horizontal distance traveled by the helicopter during this time
While the packet is falling, the helicopter continues to move horizontally. The horizontal distance \( x \) that the helicopter travels during the time \( t \) is given by:
\[ x = V \cdot t \]
Substituting the expression for \( t \):
\[ x = V \cdot \sqrt{\frac{2H}{g}} \]

### Step 4: Determine the total distance from the man when the packet is dropped
The total distance \( D \) from the helicopter to the man when the packet is dropped can be found using the Pythagorean theorem, since the horizontal distance \( x \) and the vertical distance \( H \) form a right triangle:
\[ D = \sqrt{x^2 + H^2} \]
Substituting the expression for \( x \):
\[ D = \sqrt{\left(V \cdot \sqrt{\frac{2H}{g}}\right)^2 + H^2} \]
Simplifying this:
\[ D = \sqrt{\frac{2V^2H}{g} + H^2} \]

### Final Expression
Thus, the distance of the helicopter from the man when the food packet is dropped is:
\[ D = \sqrt{H^2 + \frac{2V^2H}{g}} \]