The height of a helicopter above the ground is given by h = 2.55t3, where h is in meters and t is in seconds. At t = 1.95 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?
The helicopter's upward velocity is v=7.65 t^2, so to find out how long it takes to fall (h=0), solve
2.55 * 1.95^3 + 7.65 * 1.95^2 - 4.9t^2 = 0