# Homework Help: Kinematics Equation - Simple yet tough

1. Sep 23, 2010

### Saterial

1. The problem statement, all variables and given/known data
The height of a helicopter above the ground is given by h=3.20t^3, where h is in meters and t is in seconds. After 2.05s, the helicopter releases a small mailbag. How long after it's release does the mailbag reach the ground?

2. Relevant equations
Δd=v1Δt + 1/2aΔt^2

3. The attempt at a solution
At first when I looked at this question, I thought it was really easy, however it didn't work out well.

Δd=3.20t^3
V1=0
a=9.8m/s^2

Δd=v1Δt+1/2aΔt^2
3.20t^3=-4.9t^2

My approach to this is most likely wrong but I can't figure out why. I picked an arbitrary direction of down as positive and my thoughts on this question are, can't I simply divide 3.20t^3 by 4.9t^2 to cancel out the t leaving you with just t? However, that didn't work. Additionally, I assumed that the 2.05s is not used, but if there is, I can't see how it would be of relevance to the question.

2. Sep 23, 2010

### Forty-Two

You have the right idea about which equations to use.

Your definitions of your t's is not correct however. The t in the equation, h=3.20t^3, is being used to refer to the height of the helicopter at a certain time t.

What you are trying to solve for however is a different time, the time it takes the bag to fall to the ground. These two times are not necessarily equal.

Hope this helps and let me know if you need more guidance.

3. Sep 23, 2010

### Saterial

Is there a purpose of the 2.05s ? I don't see how it's of any relevance to the time it takes for the mailbag to fall down.

4. Sep 23, 2010

### hogrampage

Yes, that can be used to find the height at which the helicopter is when the bag is launched. This will be your y value (the height it must fall). Use that to solve for t using:

y = 1/2gt^2

(Note: The actual equation is y-yo = vosin(theta)t - 1/2gt^2, but it is a launch of 0 degrees, so sin(0) = 0. And, I just replaced -1/2gt^2 with 1/2gt^2 because the bag is falling downward - in the direction of gravity).

Someone correct me if I'm wrong.