A small mailbag is released from a helicopter that is descending steadily at 1.67 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 1.67 m/s? V = m/s d =

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter2: Motion In One Dimension
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A small mailbag is released from a helicopter that is descending steadily at 1.67 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 1.67 m/s?
V =
m/s
d =
Transcribed Image Text:A small mailbag is released from a helicopter that is descending steadily at 1.67 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 1.67 m/s? V = m/s d =
Expert Solution
Step 1

there are three equatio  of motion 

v = u + at 

v- u2 = 2as 

s = ut + 0.5 gt2 

here u = initial velocity 

v = final velocity 

a = acceleration of body 

t = time 

s = displacement or of body 

 

acceleration due to gravity is given by g = 9.8 m/s2 and is in downward direction  

if body has constant speed v then distance moved by body in t sec is d 

d = vt 

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