The height of a helicopter above the ground is given by h = 2.55t^3, where h is in meters and t is in seconds. At t = 1.65 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?
The above equation, and d = 1/2at (because initial velocity is zero)
The Attempt at a Solution
2.55(1.65)^3 = 11.45491875 m/s
11.45491875 = 1/2 (-9.8) (t^2)
22.9098375 = (-9.8) (t^2)
-2.3377 = t^2
You then get an imaginary number, but regardless, because time cannot be negative, I pretended like the answer was positive and took the square root.
I got uhhh 1.528966488, but when I plug it into the online homework, it's incorrect?