Helicopter problem

1. Feb 28, 2008

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

2. Feb 29, 2008

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.

3. Feb 29, 2008

physixguru

WHO IN THE WORLD TOLD YOU THAT THE INITIAL SPEED OF THE BAG IS ZERO???

4. Feb 29, 2008

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

Last edited: Feb 29, 2008