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 = 1.60 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Answer 1

h = 2.55t^3 = 2.55(1.60)^3 = 10.44 m.

h = Vo*t + 0.5g*t^2 = 10.44 m,
0 + 4.9t^2 = 10.44,
t^2 = 2.13,
t = 1.46 s.