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A small mailbag is released from a helicopter that is descending steadily at 2.82 m/s.(a) After 3.00 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 2.82 m/s?v = m/sd = m

User Fibericon
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a)

When the package is released from the moving helicopter, the package and the helicopter has a common speed. The package is in freefall. We would calculate the speed of the helicopter after a given time t by applying the formula,

v = vo + gt

where

vo is the initial velocity of of the package and it is equal to the speed of the helicopter

v is the final velocity of the package after time t

g is th acceleration due to gravity

From the information given

vo = 2.82

t = 3

g = 9.8

Thus,

v = 2.82 + 9.8 * 3 = 2.82 + 29.4

v = 32.22 m/s

After 3.00 s, the speed of the mailbag is 32.22 m/s

b) We want to calculate the distance covered by the mailbag in 3 s. We would apply the formula which is expressed as

s = vot + 1/2 x g x t^2

where

s is the distance

vo = 2.82

g = 9.8

t = 3

Thus,

s = 2.82 x 3 + 1/2 x 9.8 x 3^2 = 8.46 + 44.1

s = 52.56 m

Since we want to calculate the distance from the helicopter, we would calculate the diatance that the helicopter also travelled downwards in 3 s by applying the formula for calculating distance which is expressed as

distance = speed x time

Thus

distance = 2.82 x 3 = 8.46 m

Difference in distance = 52.56 - 8.46 = 44.1

The package is 44.1 m from the helicopter

c) If the helicopter is moving upwards, it would be thrown out and it would attain a certain height before it starts descending. The height is calculated by the formula,

h = vo^2/2g

By substituting the values,

h = 2.82^2/2 x 9.8 = 0.406 m

When the mail bag attains this height, it will start moving downwards. At this height, the final velocity is zero. We would calculate the time taken to attain this height by applying the formula,

v = vo - gt

v = 0

Thus,

0 = 2.82 - 9.8 x t

9.8t = 2.82

t = 2.82/9.8 = 0.288

The time left for freefall within the first 3 seconds is

3 - 0.288 = 2.712 s

The height attained by the mailbag in 2.712s is calculated by the formula,

h = gt^2/2

h = 9.8 x 2.712^2/2 = 36.04 m

Distance travelled by helicopter by ascending upward in 3 s is

distance = 2.82 x 3 = 8.46

Height of mailbag from final position after 3 seconds is

36.04 - 0.406 = 35.634

Difference in distance = 35.634 + 8.46 = 44.094

The package is 44.094 m from the helicopter

For the velocity of the mailbag after 3 seconds,

v = - vo + gt

v = - 2.82 + 9.8 x 3 = - 2.82 + 29.4

v = 26.54 m/s

User JTG
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