a food packet is released from a helicopter which is rising steadily at 2 m/s

after 2 sec ( i ) what is the velocity of the packet ( ii) how far is it below the helicopter?

Given

t= 2sec

u= 2 sec

a= -9.8

i)v = u + at = 2 + -9.8 x 2 = 17.6 m/s downwards.

ii) Again from the equation of motion

 s= ut+(1/2)at²

= 2x2 + (1/2) x - 9.8 x 2² = 15.6 m

The helicopter rises at rate of 2m/s so after 2 sec it will have risen

 i.e. d= v x t = 2x2 = 4 m

Therefore their separation will be  or the food packet is = 15.6 + 4 =19.6 m