A food packet is released from a helicopter which is trising steadily at ` 3 ms^(-1)`. After ` 3 seconds`, (i) what is the velocity of the packet? (ii) how far is it below the helicopter ? Take ` g= 9.8 m//s^2`.

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Given Data:
- Velocity of helicopter (upward) = \(3 \, \text{m/s}\)
- Time after release = \(3 \, \text{s}\)
- Acceleration due to gravity, \(g = 9.8 \, \text{m/s}^2\)

### Part (i): Finding the velocity of the packet after 3 seconds
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