Asmall mailbag is released from a helicopter that is descending steadily at 1.39 m/s. (a) after 5.00 s, what is the speed of the mailbag? (b) how far is it below the helicopter? (c) what are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.39 m/s?

a) 50.4 m/s

b) 122.5 m

c) 47.61 m/s , 122.5 m

Explanation:

a)

v₀ = initial velocity of the small mailbag = - 1.39 m/s

a = acceleration = - 9.8 m/s²

v = final velocity after "5 sec"

t = time interval = 5 sec

using the equation

v = v₀ + a t

v = - 1.39 + (- 9.8) (5)

v = - 50.4 m/s

speed of mailbag after "5 sec" is 50.4 m/s

b)

D = distance traveled by helicopter in "5 sec "

d = distance traveled by small mailbag in "5 sec"

distance traveled by helicopter in "5 sec " is given as

D = v₀ t

D = (- 1.39) (5)

D = - 6.95 m

distance traveled by small mailbag in "5 sec " is given as

d = v₀ t + (0.5) a t²

d = (- 1.39) (5) + (0.5) (- 9.8) (5)²

d = - 129.45 m

y = distance below the helicopter

distance below helicopter is given as

y = |d - D| = |(- 129.45) - (- 6.95)| = 122.5 m

c)

v₀ = initial velocity of the small mailbag = 1.39 m/s

a = acceleration = - 9.8 m/s²

v = final velocity after "5 sec"

t = time interval = 5 sec

using the equation

v = v₀ + a t

v = 1.39 + (- 9.8) (5)

v = - 47.61 m/s

speed of mailbag after "5 sec" is 47.61 m/s

D = distance traveled by helicopter in "5 sec "

d = distance traveled by small mailbag in "5 sec"

distance traveled by helicopter in "5 sec " is given as

D = v₀ t

D = (1.39) (5)

D = 6.95 m

distance traveled by small mailbag in "5 sec " is given as

d = v₀ t + (0.5) a t²

d = (1.39) (5) + (0.5) (- 9.8) (5)²

d = - 115.55 m

y = distance below the helicopter

distance below helicopter is given as

y = |d - D| = |(- 115.55) - (6.95)| = 122.5 m

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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