How Should Gravity Be Signed in Kinematic Equations?

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Homework Help Overview

The problem involves a package dropped from a helicopter moving upwards, requiring the application of kinematic equations to determine the height from which it was released. The context includes considerations of directionality in velocity and acceleration due to gravity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply a kinematic equation but questions the sign convention for gravity and the implications of their chosen coordinate system. Other participants clarify the need for consistency in sign usage and the interpretation of results based on the defined origin.

Discussion Status

Participants have provided guidance on maintaining consistent sign conventions and interpreting the results based on the chosen reference point. The original poster expresses understanding after receiving clarification, indicating a productive direction in the discussion.

Contextual Notes

There is an emphasis on the importance of defining the coordinate system and the implications of the signs used for velocity and acceleration in the context of the problem.

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Homework Statement



A package is dropped from a helicopter moving upwards at 15 m/. If it takes 25.0s before the package hits the ground, how high above the ground was the package when it was released. Disregard air resistance.

Homework Equations



I used this kinematic equation

s_f=s_i +v_i t + 1/2 a (t)^2

The Attempt at a Solution



So I set it up with the initial velocity being +15m/s since it is going up and the gravity as -9.80m/s^2 as it is going down.

Here is my question. Is that right? I am completely lost when it comes to g being positive or negative. I understand you use whichever sign coresponds to your setup directions. But in this case I end up with -2688. Do I just disregard the sign on the distance? Or is could it be negative because my set up states the package was released on the x-axis so technically the end position is negative because it's below the axis? So in that case I do change the sign?

I figure this is pretty important to figure out as it can have a major effect on the outcome of these problems.
 
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If your velocity is +, then g must be negative in that equation. You shouldn't be getting a negative number though. When you do vt-1/2at^2 you get a negative number, but remember that here, your s_f is 0 and you're solving for s_i not the other way around (so you have to bring everything to the other side).
 
Using +15 m/s is right if you specify "up" as the positive direction. Once you do that, you must be consistent with the other vector quantities. You must use a = - 9.8 m/s2 because the acceleration of gravity is "down".

It is also important to specify your origin with respect to which you measure distances. If you say si = 0, then your origin is the initial position of the package and you should expect sf to be a negative number (below the origin). If you say sf = 0 (i.e. you assume that your origin is the ground) and solve for si, then you should expect a positive number (above ground).

You should not change any sign, but you should interpret the sign that you get on the basis of your assumptions about which way is "up" and where your origin is.
 
Ahhhhh I get it! :bugeye:

And now that your two explained it and I read the question again, I see why my answer was negative but the actual answer isn't.

I found from the helo down, which is negative 2688, but I need from the ground to the helo which is positive once I switch things around.

I thought I had it but then did this problem and was lost again.

Now that I have these explanations, I understand it 100%.

Thank you
 

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