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.