What is the acceleration of a falling object from a balloon?

[Eclair_de_XII]

I think here, instead of introducing spurious negative signs, it would be better to use g = -9.8 m/s2 along with the distance fallen of -87.347 m.
v02 still comes out positive.

But won't it be imaginary since $v_0$ is still negative?

The change in position will also be negative since the package was 87.347 m above the ground before it fell.

But then you'd have to move the negative displacement to the other side, making it positive.

 

[Eclair_de_XII]

Well, anyway, thank you so much for everything. You and everyone have been a big help to me with my self-study. Without this forum, I don't believe I would be able to learn physics on my own. So... thank you. :)

 

[SteamKing]

But won't it be imaginary since $v_0$ is still negative?

The equation is v² = u² + 2 a s

There is no negative sign on the velocity terms. The product of -9.8 and -87.347 is also positive.

But then you'd have to move the negative displacement to the other side, making it positive.

Not if you take x = 0 being the ground and x₀ = 87.347 m as the height from which the package started falling. Here, v₀ = 0 and g = -9.8 m/s², making the equation:

x - x₀ = v₀ t + (1/2)*g*t² → 0 - 87.347 = (1/2) * (-9.8) * t²