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

The vertical velocity is
V = dh/dt = 7.65 t^2 (from differential calculus)
At t = 1.60 s, the height is
h = 10.45 m
and the velocity is
V = 19.58 m/s

Now let t be measured from the time of release. The mailbag reaches the ground when

H = 10.45 + 19.58 t - (g/2)t^2 = 0

Solve for t.