A small mailbag is released from a helicopter that is descending steadily at 3.00 m/s.
(a) After 2.00 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 3.00 m/s?
v = m/s
d = m
To solve this problem, we need to use the equations of motion.
(a) After 2.00 s, what is the speed of the mailbag?
To find the speed of the mailbag, we can use the equation:
v = u + at
Where:
v = final velocity (speed)
u = initial velocity (speed)
a = acceleration
t = time
In this case:
u = 0 m/s (since the mailbag is released from rest)
a = 3.00 m/s^2 (acceleration due to the descending helicopter)
t = 2.00 s
Plugging in these values into the equation, we get:
v = 0 + (3.00 m/s^2)(2.00 s)
v = 0 + 6.00 m/s
v = 6.00 m/s
Therefore, the speed of the mailbag after 2.00 s is 6.00 m/s.
(b) How far is it below the helicopter?
To find the distance the mailbag has traveled below the helicopter, we can use the equation:
d = ut + (1/2)at^2
Where:
d = distance
u = initial velocity (speed)
t = time
a = acceleration
In this case:
u = 0 m/s (since the mailbag is released from rest)
a = 3.00 m/s^2 (acceleration due to the descending helicopter)
t = 2.00 s
Plugging in these values into the equation, we get:
d = 0(2.00 s) + (1/2)(3.00 m/s^2)(2.00 s)^2
d = 0 + (1/2)(3.00 m/s^2)(4.00 s^2)
d = (1/2)(3.00 m/s^2)(4.00 s^2)
d = (1/2)(3.00)(4.00) m
d = 6.00 m
Therefore, the mailbag is 6.00 meters below the helicopter.
(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 3.00 m/s?
If the helicopter is rising steadily at 3.00 m/s, the answers to parts (a) and (b) will be different.
(a) After 2.00 s, what is the speed of the mailbag?
To find the speed of the mailbag in this case, we can again use the equation:
v = u + at
Where:
v = final velocity (speed)
u = initial velocity (speed)
a = acceleration
t = time
In this case:
u = 0 m/s (since the mailbag is released from rest)
a = -3.00 m/s^2 (acceleration due to the rising helicopter)
t = 2.00 s
Plugging in these values into the equation, we get:
v = 0 + (-3.00 m/s^2)(2.00 s)
v = 0 - 6.00 m/s
v = -6.00 m/s
Therefore, the speed of the mailbag after 2.00 s is -6.00 m/s. Note that the negative sign indicates that the mailbag is moving upwards.
(b) How far is it below the helicopter?
To find the distance the mailbag has traveled below the helicopter, we can use the same equation as before:
d = ut + (1/2)at^2
Where:
d = distance
u = initial velocity (speed)
t = time
a = acceleration
In this case:
u = 0 m/s (since the mailbag is released from rest)
a = -3.00 m/s^2 (acceleration due to the rising helicopter)
t = 2.00 s
Plugging in these values into the equation, we get:
d = 0(2.00 s) + (1/2)(-3.00 m/s^2)(2.00 s)^2
d = 0 + (1/2)(-3.00 m/s^2)(4.00 s^2)
d = (1/2)(-3.00 m/s^2)(4.00 s^2)
d = (1/2)(-3.00)(4.00) m
d = -6.00 m
Therefore, the mailbag is 6.00 meters above the helicopter. Again, note the negative sign, which indicates that the mailbag is below the helicopter.