Question

A food packet is released from a helicopter rising steadily at 2m/s. After 2 sec: (i)what is the velocity of packet?(ii) how far is it below the helicopter.g=9.8m/s

Answer 1

Hello!!

We know that Velocity=gt+u

=9.8*2+2

=21.6m/s

So velocity after 2s is 21.6 m/s

and

Distance=ut+1/2at^2

=4+1/2*9.8*2^2

=4+19.6

=23.6m
Thanks!!
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