A small mailbag is released from a helicopter that is descending steadily at 1.9
ID: 1877787 • Letter: A
Question
A small mailbag is released from a helicopter that is descending steadily at 1.91 m/s (a) After 5.00 s, what is the speed of the mailbag? V-50.91 m/s (b) How far is it below the helicopter? d 26.41X Your response differs from the correct answer by more than 10%. Double check your calculations. m (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.91 m/s? v# 47.09 m/s d 112.95X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation results to at least four-digit accuracy to minimize roundoff error. m am out all intermediateExplanation / Answer
take downward direction as negative
a)
vi = -1.91 m/s
a = -9.8 m/s^2
t = 5.00 s
use:
vf = vi + a*t
= -1.91 + (-9.8)*5.00
= -1.91 - 49
= -50.91 m/s
Answer: 50.91 m/s
positive because its asking speed
b)
vf^2 = vi^2 + 2*a*d
50.91^2 = 1.91^2 + 2*(-9.8)*d
-19.6*d = 2588.2
d = —132 m
Answer: 132 m
positive because its asking distance
c)
vi = 1.91 m/s
a = -9.8 m/s^2
t = 5.00 s
use:
vf = vi + a*t
= 1.91 + (-9.8)*5.00
= 1.91 - 49
= -47.09 m/s
Answer: 47.09 m/s
positive because its asking speed
vf^2 = vi^2 + 2*a*d
47.09^2 = 1.91^2 + 2*(-9.8)*d
-19.6*d = 2213.82
d = —113 m
Answer: 113 m
positive because its asking distance
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