๐Ÿš A Helicopter Drops a Food Packet – Motion in a Straight Line | JEE Physics | Doubtify JEE


๐Ÿ“Œ Question:

A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height of h. The time taken by the packet to reach the ground is close to:
(Take g as the acceleration due to gravity)

A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height of h. The time taken by the packet to reach the ground is close to: (Take g as the acceleration due to gravity)


๐Ÿ“ท Solution Image:

A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height of h. The time taken by the packet to reach the ground is close to: (Take g as the acceleration due to gravity)


A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height of h. The time taken by the packet to reach the ground is close to: (Take g as the acceleration due to gravity)


A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height of h. The time taken by the packet to reach the ground is close to: (Take g as the acceleration due to gravity)


▶️ Watch Full Video Explanation:



๐Ÿ“š Concept Breakdown:

This question is a classic case of motion in a straight line under gravity, with a twist — the object is not stationary when dropped. The helicopter is accelerating upwards, which means the packet also gains some initial velocity upward at the moment it is dropped.

To solve this, we must:

  1. Calculate the initial velocity of the packet (since it accelerates with g until height h).

  2. Apply the equations of motion under gravity to find the total time taken for the packet to hit the ground.

We use:
u = gt (Initial upward velocity when released)
s = h + ½gt² (Total height from which it falls, since the chopper was accelerating)
Then apply the kinematic equation:
s = ut + ½gt²
But here gravity is pulling down, so acceleration becomes –g.

Finally, solving for t gives us the total time taken by the packet to reach the ground.


๐Ÿ’ก Why This Question is Important:

✅ This is a frequently asked JEE Main level conceptual question.
✅ It mixes kinematics with real-life situations – great for understanding motion in vertical direction.
✅ Helps in mastering concepts of non-zero initial velocity and acceleration due to gravity in reverse direction.


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๐Ÿ“Œ Stay tuned on Doubtify JEE for more such real-world inspired JEE problems explained in a simple and relatable way!


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