A Small Mailbag Is Released From A Helicopter That Is Descen
A Small Mailbag Is Released From a Helicopter That Is Descending Stead
A small mailbag is released from a helicopter that is descending steadily at 2.42 m/s. (a) After 5.00 s, what is the speed of the mailbag? v = 1 -61.1 The response you submitted has the wrong sign. [removed][removed] m/s (b) How far is it below the helicopter? d = 2 [removed] The response you submitted has the wrong sign. [removed][removed] m (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.42 m/s? v = 3 [removed] The correct answer is not zero. [removed][removed] m/s d = 4 [removed] The correct answer is not zero. [removed][removed] m
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The problem involves analyzing the motion of a mailbag released from a helicopter moving with a constant velocity either downward or upward. Since the helicopter moves steadily at a constant velocity, the mailbag's initial velocity relative to the ground is the same as that of the helicopter—either downward at 2.42 m/s or upward at 2.42 m/s. The main goal is to determine the velocity of the mailbag after a certain period and the displacement relative to the helicopter, considering only gravity's influence after release.
Part (a): Mailbag Released from a Descending Helicopter
When the helicopter is descending steadily at 2.42 m/s, the initial velocity (v0) of the mailbag relative to the ground is -2.42 m/s (taking downward as negative). After its release, the mailbag will be accelerated by gravity, which is approximately 9.81 m/s2 downward.
The velocity of the mailbag after time t can be calculated using the equation:
V = V0 + g * t
where:
- V0 = -2.42 m/s
- g = 9.81 m/s2
- t = 5.00 s
Plugging in the values:
V = -2.42 m/s + 9.81 m/s2 * 5.00 s = -2.42 m/s + 49.05 m/s = 46.63 m/s
The positive sign indicates that the mailbag's velocity is downward at 46.63 m/s after 5 seconds.
Part (b): Displacement of the Mailbag Below the Helicopter
The displacement relative to the initial position is given by:
d = V0 t + 0.5 g * t2
Substituting the known values:
d = -2.42 m/s 5.00 s + 0.5 9.81 m/s2 * (5.00 s)2
d = -12.10 m + 0.5 9.81 m/s2 25 s2
d = -12.10 m + 0.5 9.81 m/s2 25
d = -12.10 m + 122.625 m = 110.525 m
The positive displacement indicates that the mailbag is approximately 110.53 meters below the initial release point (the helicopter's position). The negative initial velocity and positive displacement together confirm the mailbag is falling downward relative to the initial point.
Part (c): If the Helicopter is Rising at 2.42 m/s
In this case, the initial velocity of the mailbag is +2.42 m/s (upward). The acceleration due to gravity still acts downward at 9.81 m/s2. The equations are similar, but initial velocity signs are adjusted accordingly.
Velocity after 5 seconds:
V = V0 + g t = +2.42 m/s - 9.81 m/s2 5.00 s = +2.42 m/s - 49.05 m/s = -46.63 m/s
The negative value indicates that after 5 seconds, the mailbag's velocity is downward at 46.63 m/s relative to the ground, meaning it has overtaken the upward motion of the helicopter and begun falling downward.
The displacement after 5 seconds:
d = V0 t + 0.5 g t2 = +2.42 m/s 5.00 s + 0.5 (-9.81) m/s2 25
d = 12.10 m - 122.625 m = -110.525 m
The negative displacement indicates the mailbag is approximately 110.53 meters below the initial release point, matching the earlier calculations but in the upward direction initially and then falling downward as time progresses.
Summary
In both cases, the timing and initial velocities significantly influence the mailbag's motion. The key takeaways are that gravity accelerates the mailbag downward regardless of the initial movement of the helicopter, causing it to reach high downward velocities over time. The signs of velocity and displacement depend on whether the helicopter is descending or ascending, but the magnitude of displacement after 5 seconds remains approximately 110.5 meters in both scenarios, illustrating the independence of displacement magnitude from initial velocity when gravity acts over time.
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