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?
u= 2 sec
i)v = u + at = 2 + -9.8 x 2 = 17.6 m/s downwards.
ii) Again from the equation of motion
= 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