<u>Answer:</u>
a) Speed of mailbag after 3 seconds = 32.4 m/s
b) Package is 44.1 meter below helicopter
c) If the helicopter was rising steadily at 3.00 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Package is 44.1 meter below helicopter
<u>Explanation:</u>
a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Initial velocity = 3 m/s, acceleration = 9.8 
and time = 3 seconds.
v = 3+9.8*3 = 32.4 m/s
Speed of mailbag after 3 seconds = 32.4 m/s
b) We have equation of motion , 
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 53.1 meter.
So package is (53.1-9)meter below helicopter = 44.1 m
c) Initial velocity = -3 m/s, acceleration = 9.8 and time = 3 seconds.
v = -3+9.8*3 = 26.4 m/s
Speed of mailbag after 3 seconds = 26.4 m/s
Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 .
Distance traveled by helicopter = 9 meter.
Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 .
Distance traveled by package = 35.1 meter.
So package is (35.1+9)meter below helicopter = 44.1 m
