Mailbag dropped from helicopter moving at 2.98 m/s

[nbroyle1]

A small mailbag is released from a helicopter that is descending steadily at 2.98 m/s.

(a) After 5.00 s, what is the speed of the mailbag?
(b) How far is it below the helicopter?
(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.98 m/s?

I tried using the equation v(final)=v(initial)+gt and ended up with the wrong answer.
v(final) is what I'm solving for and I plugged in 2.98 as the initial velocity, 5s for time and -9.8 for gravity. I must be using the wrong equation could I get some help?

 

[Infinitum]

Hi nbroyle!

I tried using the equation v(final)=v(initial)+gt and ended up with the wrong answer.
v(final) is what I'm solving for and I plugged in 2.98 as the initial velocity, 5s for time and -9.8 for gravity. I must be using the wrong equation could I get some help?

Remember which direction you have considered as positive. If it is downwards, the initial velocity is downwards, and so is g. How will get your equations now? If positive upwards, both the initial velocity and g will be negative.

 

[nbroyle1]

I don't quite understand, what do you mean by how will I get my equations?

 

[azizlwl]

Check your convention of direction.
If you put acceleration negative, it means all motions downward negative.
Take downward as positive for easier calculation.

 

[nbroyle1]

Ok so If I chose the positive means downward convention then the equation would read:
v=(9.8)(5)+2.98 correct?

 

[nbroyle1]

It is saying the answer is incorrect when I calculate the velocity like this as well.

 

[nbroyle1]

nevermind calculation error oops thanks for the help