Post a New Question

physics

posted by .

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.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question