1. The problem statement, all variables and given/known data The height of a helicopter above the ground is given by h = 3.30t3, where h is in meters and t is in seconds. After 1.80 s, the helicopter releases a small mailbag. Assume the upward direction is positive and the downward direction is negative. Already solved for Initial velocity for when mailbag's released, V0=32.08 m/s and also solved for initial height at 19.2 m and maximum height at 71.8 m. Acceleration of gravity = -9.81 m/s/s The actual question I have is What is the velocity of the mailbag when it hits the ground? 2. Relevant equations Vxf2 = Vxi2 + 2(ax)(xf - xi) 3. The attempt at a solution I plugged in 32.08 = Vxi , Ax = 9.81 and delta x(xf-xi) = 52.6 and I got 45.39 after taking the square root, which I know is wrong because I got the real answer by accident, the real answer being 37.96 m/s. I got the real answer by not squaring 32.08 and having delta x being 71.8. So basically this: Vxf = sqrt(2(9.81)(71.8)+32.08) When I was looking over at YahooAnswers, I found that the delta x was something like -19.2, and acceleration is of course -9.81, so I plugged that in and got 37.49 m/s, which is extremely similar to my accident answer. So basically my query is this: was my delta x wrong? Was I supposed to subtract 0 and 19.2 and use that as my delta x? Just want to know how to do this type of problem correctly instead of accidentally stumbling on it.