Question

A helicopter is flying horizontally with a speed 'v' at an altitude 'h' 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 ?

• A. \(\sqrt{\frac{2gh}{v^2} + h^2}\)
• B. \(\sqrt{\frac{2v^2h}{g} + h^2}\)
• C. \(\sqrt{2ghv^2 + h^2}\)
• D. \(\sqrt{\frac{2ghv^2 + 1}{h^2}}\)

Hint:

Always break projectile motion into independent horizontal and vertical components. Time is the common variable connecting them.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When the packet is dropped, it has an initial horizontal velocity \( v \) and zero initial vertical velocity. The time it takes to reach the ground depends only on the altitude \( h \).
Step 2: Key Formula or Approach:
Time of flight \( t = \sqrt{\frac{2h}{g}} \).
Horizontal range \( R = v \times t = v \sqrt{\frac{2h}{g}} \).
Step 3: Detailed Explanation:
The helicopter is at a height \( h \). The man is at the point where the packet will land, which is at a horizontal distance \( R \) from the point directly below the helicopter at the moment of release.
The distance \( D \) between the helicopter and the man at the moment of drop is the hypotenuse of the right triangle formed by \( h \) and \( R \):
\[ D = \sqrt{h^2 + R^2} \] Substitute \( R \):
\[ D = \sqrt{h^2 + \left( v \sqrt{\frac{2h}{g}} \right)^2} \] \[ D = \sqrt{h^2 + \frac{2v^2h}{g}} \] Step 4: Final Answer:
The distance is \(\sqrt{\frac{2v^2h}{g} + h^2}\).