Question

The height of a helicopter above the ground is given by h = 2.80t3, where h is in meters and t is in seconds. At t = 1.55 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? ...?

Answer 1

Do this.

H = 2.80t^3

h = 2.80 (1.55)^3 = 10.43

10.43 = 1/2 gt^2

10.43 x 2/9.8  = t^2

t = √2.12

= (Answer is 1.45)

Hope this helps!