A small mailbag is released from a helicopter that is descending steadily at 2.58 m/s.
(a) After 3.00 s, what is the speed of the mailbag?
v =
32.01
Correct: Your answer is correct.
m/s
(b) How far is it below the helicopter?
d =
41.9 ?????????????/
Incorrect: Your answer is incorrect.
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m
(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.58 m/s?
v =
26.85
Correct: Your answer is correct.
m/s
d =
43.825
Correct: Your answer is correct.
m
a. V = Vo + g*t = 2.58 + 9.8*3 = 31.98 m/s.
b. V^2 = Vo + 2g*d = 31.98^2,
2.58 + 19.6d = 1022.72,
d = 52. m.
c. a. V = -2.58 + 9.8*3 = 26.82 m/s. b. -2.58 + 19.6*d = 1022.72.
the moving helicopter is an inertial reference frame
... you used that in (a) , what about (b)?
(b) the bag is accelerating away from the helicopter
... acceleration is g , there is no initial velocity (relative to the helicopter)
... distance = 1/2 * g * 3^2
To solve this problem, we need to break it down into parts.
(a) To find the speed of the mailbag after 3.00 s, we can use the formula:
v = v0 + at
where:
v is the final speed,
v0 is the initial speed,
a is the acceleration, and
t is the time.
In this case, the helicopter is descending steadily at 2.58 m/s, so the initial speed (v0) is 0 m/s and the acceleration (a) is -2.58 m/s^2 (negative because it's descending).
Plugging in the values, we have:
v = 0 + (-2.58 * 3.00)
v = -7.74 m/s
However, we only want the magnitude of the speed, so we take the absolute value:
v = |-7.74| = 7.74 m/s
Therefore, the speed of the mailbag after 3.00 s is 7.74 m/s.
(b) To find how far below the helicopter the mailbag is after 3.00 s, we can use the formula:
d = v0t + (1/2)at^2
where:
d is the distance,
v0 is the initial speed,
a is the acceleration, and
t is the time.
In this case, the initial speed (v0) is also 0 m/s and the acceleration (a) is -2.58 m/s^2.
Plugging in the values, we have:
d = 0 * 3.00 + (1/2)(-2.58)(3.00^2)
d = 0 + (1/2)(-2.58)(9.00)
d = 0 + (-3.87)
d = -3.87 m
However, we only want the magnitude of the distance (since it's below the helicopter), so we take the absolute value:
d = |-3.87| = 3.87 m
Therefore, the mailbag is 3.87 m below the helicopter after 3.00 s.
(c) In this case, the helicopter is rising steadily at 2.58 m/s, so the acceleration (a) is now positive since it's ascending.
Using the same formulas as before, we find:
v = v0 + at
v = 0 + 2.58 * 3.00
v = 7.74 m/s
Therefore, the speed of the mailbag after 3.00 s is 7.74 m/s.
d = v0t + (1/2)at^2
d = 0 * 3.00 + (1/2)(2.58)(3.00^2)
d = 0 + (1/2)(2.58)(9.00)
d = 0 + 11.61
d = 11.61 m
Therefore, the mailbag is 11.61 m below the helicopter after 3.00 s.