The height of a helicopter above the ground is given by h = 2.90t3, where h is in meters and t is in seconds. At t = 1.75 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Answer 1

Using standard equation the height is given by:

s=uT+0.5aT^2

if it is dropped u=0

s=0.5aT^2

2.90 t^3 = 0.5 9.8 T^2

substitute value for t as 1.75 s

and find T

Answer 2

tried your way and it didn't work,