# physics

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A small mailbag is released from a helicopter that is descending steadily at 1.01 m/s.
(a) After 4.00 s, what is the speed of the mailbag?
v = m/s

(b) How far is it below the helicopter?
d = m

(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.01 m/s?
v = m/s
d = m

• physics -

(a) v=vₒ+gt = 1.01+9.8•4=40.21 m/s

(b) the distance covered by mailbag (downward)
s1= vₒt+gt²/2 =1.01•4+9.8•4²/2=82.44 m
the distance covered by the helicopter (downward)
s2=v•t =1.01•4= 4.04 m
Δs=s1-s2= 82.44 – 4.04=78.4m
(c) Upward motion of the mailbag
h1=v²/2g =1.01²/2•9.8=0.053 m.
This motion takes the time
v=vₒ-g•t1.
v=0
t1= vₒ/g=1.01/9.8 =0.1 s
Free fall during
t2=4-0.1 =3.9 s
h2 = g•t ²/2 = 9.8•3.9²/2 =74 m
Upward motion of the helicopter
s2=v•t =1.01•4= 4.04 m
Position of the mailbag
74-0.053 =73.947 m

Δs=73.947+4.04=77.987m.

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