(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

The height of a helicopter above the ground is given by h=3.20t^3, where h is in meters and t is in seconds. After 2.05s, the helicopter releases a small mailbag. How long after it's release does the mailbag reach the ground?

2. Relevant equations

Δd=v1Δt + 1/2aΔt^2

3. The attempt at a solution

At first when I looked at this question, I thought it was really easy, however it didn't work out well.

Δd=3.20t^3

V1=0

a=9.8m/s^2

Δd=v1Δt+1/2aΔt^2

3.20t^3=-4.9t^2

My approach to this is most likely wrong but I can't figure out why. I picked an arbitrary direction of down as positive and my thoughts on this question are, can't I simply divide 3.20t^3 by 4.9t^2 to cancel out the t leaving you with just t? However, that didn't work. Additionally, I assumed that the 2.05s is not used, but if there is, I can't see how it would be of relevance to the question.

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# Homework Help: Kinematics Equation - Simple yet tough

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