A small mailbag is released from a helicopter that is descending steadily at 2.20 m/s.
How far is it below the helicopter?
D=
What if the helicopter is rising steadily at 2.20m/s?
D=
To find the distance below the helicopter, we can use the formula:
D = (V * t) + (0.5 * a * t^2)
where:
D is the distance below the helicopter,
V is the initial velocity of the mailbag,
t is the time for which the mailbag has been released, and
a is the acceleration of the helicopter.
Given that the helicopter is descending or rising steadily at a velocity of 2.20 m/s, we can assume that the acceleration is 0 m/s^2. Therefore, the first term in the formula becomes zero.
1. When the helicopter is descending steadily at 2.20 m/s:
In this case, the initial velocity (V) of the mailbag would also be 2.20 m/s. Assuming the mailbag is released for a duration of t seconds, the distance below the helicopter (D) can be calculated as follows:
D = (2.20 m/s) * t
2. When the helicopter is rising steadily at 2.20 m/s:
In this scenario, the initial velocity (V) of the mailbag would be -2.20 m/s (negative sign indicates the opposite direction). By using the same formula as above and plugging in the values, the distance below the helicopter (D) can be calculated as follows:
D = (-2.20 m/s) * t
So depending on the direction of the helicopter's motion (descending or rising), the value of 'V' in the formula will either be positive or negative.