A small mailbag is released from a helicopter that is descending steadily at 2.36 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 2.36 m/s?
I got the correct speeds for both situations, but incorrect distances (significant figures matter).
(a) v=51.4 m/s (c) v=46.6 m/s
(b)
bag falls -2.36*5 - 4.9*25 = -134.3
chopper descends -2.36*5 = -11.8
-134.3-(-11.8) = -122.5
if the chopper is rising, then
bag falls 2.36*5 - 4.9*25 = -110.7
chopper rises 11.8
-110.7 - 11.8 = -122.5
surprised the answer is the same? Think about it.
(a) To find the speed of the mailbag after 5.00 s, we need to use the equation v = v₀ + at, where v is the final velocity, v₀ is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0 m/s (since the mailbag was just released), the acceleration is -2.36 m/s² (since the helicopter is descending), and the time is 5.00 s.
Using the equation, we can calculate the speed:
v = 0 + (-2.36 m/s²)(5.00 s)
v = -11.8 m/s
However, since speed is a scalar quantity (magnitude only), we can disregard the negative sign. Therefore, the speed of the mailbag after 5.00 s is 11.8 m/s.
(b) To find how far below the helicopter the mailbag is after 5.00 s, we can use the equation y = y₀ + v₀t + (1/2)at², where y is the final vertical displacement, y₀ is the initial vertical displacement, v₀ is the initial velocity, a is the acceleration, and t is the time. In this case, the initial vertical displacement is 0 m (since both the helicopter and the mailbag start at the same height), the initial velocity is 0 m/s, the acceleration is -2.36 m/s², and the time is 5.00 s.
Using the equation, we can determine the vertical displacement:
y = 0 + 0 + (1/2)(-2.36 m/s²)(5.00 s)²
y = 0 + 0 + (1/2)(-2.36 m/s²)(25.0 s²)
y = 0 + 0 + (-29.5 m)
y = -29.5 m
Since the question asks for the distance below the helicopter, we can disregard the negative sign. Therefore, the mailbag is 29.5 m below the helicopter after 5.00 s.
(c) If the helicopter is rising steadily at 2.36 m/s, we need to change the sign of the acceleration. Using the same equations as before:
(a) v = v₀ + at
v = 0 + (2.36 m/s²)(5.00 s)
v = 11.8 m/s
(b) y = y₀ + v₀t + (1/2)at²
y = 0 + 0 + (1/2)(2.36 m/s²)(5.00 s)²
y = 0 + 0 + (1/2)(2.36 m/s²)(25.0 s²)
y = 0 + 0 + (29.5 m)
y = 29.5 m
Therefore, if the helicopter is rising steadily at 2.36 m/s, the speed of the mailbag after 5.00 s is 11.8 m/s, and it is 29.5 m below the helicopter.
To determine the answers to these questions, we can use the principles of kinematics. Let's break down each part and explain how to get the correct answers.
(a) After 5.00 s, what is the speed of the mailbag?
To find the speed of the mailbag after 5.00 seconds, we need to determine its velocity. Velocity is the rate of change of displacement over time. In this case, the mailbag is descending steadily at 2.36 m/s, and we want to find its speed after 5.00 seconds.
The equation we can use to solve for the velocity is:
v = u + at
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
Since the mailbag is descending steadily, its initial velocity u is 0 m/s (released from rest) and the acceleration a is -2.36 m/s^2 (negative because it is descending). So, plugging these values into the equation:
v = 0 + (-2.36 * 5.00)
v = -11.8 m/s
Note that the negative sign indicates the direction of the velocity, which is downwards. When we take the magnitude of the velocity, we get the speed:
|v| = |-11.8|
|v| = 11.8 m/s
Therefore, the correct speed of the mailbag after 5.00 seconds is 11.8 m/s.
(b) How far is it below the helicopter?
To find how far the mailbag is below the helicopter, we need to calculate the displacement. Displacement is the change in position of an object from its initial position to its final position. In this case, the mailbag is descending steadily, and we want to find the distance below the helicopter after 5.00 seconds.
The equation we can use to solve for the displacement is:
s = ut + (1/2)at^2
where:
s is the displacement,
u is the initial velocity,
a is the acceleration, and
t is the time.
Since the mailbag is descending steadily, its initial velocity u is 0 m/s (released from rest) and the acceleration a is -2.36 m/s^2 (negative because it is descending). So, plugging these values into the equation:
s = 0 * 5.00 + (1/2) * (-2.36) * (5.00)^2
s = -29.5 m
The negative sign indicates that the displacement is below the initial position. To get the magnitude of the displacement, we take the absolute value:
|s| = |-29.5|
|s| = 29.5 m
Therefore, the correct distance the mailbag is below the helicopter after 5.00 seconds is 29.5 m.
(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.36 m/s?
If the helicopter is rising steadily at 2.36 m/s, then the sign of the acceleration a would be positive (because it is ascending).
For part (a), we can use the same equation to find the velocity after 5.00 seconds. The initial velocity u would still be 0 m/s (since it starts from rest), and the acceleration a would be +2.36 m/s^2. Plugging these values into the equation:
v = 0 + (2.36 * 5.00)
v = 11.8 m/s
Therefore, the correct speed of the mailbag after 5.00 seconds, if the helicopter is rising steadily at 2.36 m/s, is indeed 11.8 m/s.
For part (b), we can use the same equation to find the displacement below the helicopter after 5.00 seconds. The initial velocity u would still be 0 m/s, and the acceleration a would be +2.36 m/s^2. Plugging these values into the equation:
s = 0 * 5.00 + (1/2) * (2.36) * (5.00)^2
s = 29.5 m
Therefore, the correct distance the mailbag is below the helicopter after 5.00 seconds, if the helicopter is rising steadily at 2.36 m/s, is indeed 29.5 m.
I apologize if my previous response did not provide the correct distances earlier. The correct distances are indeed 29.5 m for both scenarios, when significant figures are considered.