helicopter problem

[chocolatelover]

1. The problem statement, all variables and given/known data
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 1.85 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?


2. Relevant equations



3. The attempt at a solution

_____________________________h=2.55t^3
1.85s

h=2.55(1.85s)^3

Thank you very much

[methotrexate]

I'm fairly sure that you can make use of the formula:

h=v_{i}t+\frac{1}{2}at^2

And do it as you started:

2.55(1.85)^3=v_{i}t+\frac{1}{2}at^2

And then initial velocity is zero, and a is the gravitational constant g=9.8m/s^2:

2.55(1.85)^3=0t+\frac{1}{2}9.8t^2

And then solve for t.

[physixguru]

I'm fairly sure that you can make use of the formula:

h=v_{i}t+\frac{1}{2}at^2

And do it as you started:

2.55(1.85)^3=v_{i}t+\frac{1}{2}at^2

And then initial velocity is zero, and a is the gravitational constant g=9.8m/s^2:

2.55(1.85)^3=0t+\frac{1}{2}9.8t^2

And then solve for t.
WHO IN THE WORLD TOLD YOU THAT THE INITIAL SPEED OF THE BAG IS ZERO???

[chocolatelover]

Thank you very much

If it isn't 0, how would you solve it?

Would you do this?

velocity:
h=3.25t^3
h'=9.75t^2
h'(1.65)=9.75(1.62)^2
=26.5444

position:

h=3.25(1.65)^3
=14.6m

I would then use all of this to solve for t, right?

h=vit+1/2at^2
14.6m=26.5444m/s+1/2(-9.8m/s^2)t^2
t=1.56s

Thank you