Question

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

Answer 1

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.

Answer 2

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