Motion of an object dropped from a descending and ascending helicopter

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donking225
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1.
a) A small mail bag is released from a helicopter that is descending steadily at 3.34 m/s. After 2.95 s, what is the speed of the mailbag? The acceleration of gravity is 9.8 m/s2 . Answer in units of m/s

b) After the mailbag is dropped, the helicopter continues descending for 1 s but then stops. How far is the mailbag below the helicopter at 2.95 s? Answer in units of m

c) What would be the speed of the mailbag if the helicopter had been rising steadily at u = 3.34 m/s ? (Take down as positive.) Answer in units of m/s


2. vf = vi + at


3. I've already gotten the answer to a) which is 32.25m/s, but I do not know how to do b) or c) from my answer to a).
 
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donking225 said:
I've already gotten the answer to a) which is 32.25m/s, but I do not know how to do b) or c) from my answer to a).

For b), you need a kinematic equation (valid for constant acceleration) that involves distance (usually denoted s). Do you have any such?

c) is exactly the same as a) but with a different vi.
 
haruspex said:
For b), you need a kinematic equation (valid for constant acceleration) that involves distance (usually denoted s). Do you have any such?

c) is exactly the same as a) but with a different vi.

would this equation be appropriate for b) s=vi*t+1/2*at2
 
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For c) would I use -3.34 as the vi?
 
Thanks for the help, I've got c) correct but am still confused about b).
 
Find s for both objects, what is the difference?

Note that the helicopter is not in free fall... Descending steadily...

I am assuming that the helicopter does not undergo a constant acceleration up during the 1s. The way it is written implies to me the helicopter stops instantaneously after 1s.
 
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pgardn said:
Find s for both objects, what is the difference?

Note that the helicopter is not in free fall... Descending steadily...

I am assuming that the helicopter does not undergo a constant acceleration up during the 1s. The way it is written implies to me the helicopter stops instantaneously after 1s.

Thanks, looks like I missed the obvious. I got the correct answer now.