# Helicopter Motion Question

1. Sep 15, 2007

### Vii

1. The problem statement, all variables and given/known data

The height of a helicopter above the ground is given by $$h=3.00t^3$$, where h is in meters and t is in seconds. After 2.00s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

2. Relevant equations

$$d=v_1t + \frac{1}{2}at^2$$

3. The attempt at a solution

First, I figured out what the height would be at 2.00s by substituting t with 2.00s.

$$h=3.00(2.00)^3$$
$$h=24m$$

Then I took $$d=v_1t + \frac{1}{2}at^2$$ and plugged in 24m at d and assumed $$v_1$$ as 0.

$$24=0t + \frac{1}{2}at^2$$
$$24=\frac{1}{2}at^2$$

And plugged in 9.8m/s^2 into a.

$$24=\frac{1}{2}9.8t^2$$

Which gave me t=2.21s

The answer is wrong and I cannot figure out how to do this problem.

I think my assumption of $$v_1=0$$ is wrong. Furthermore, I think my assumption that the helicopter stops ascending at the time the package is dropped is also wrong. I've racked my brain on this question for a good 2 days now and I still can't figure out how to arrive at the answer.

2. Sep 15, 2007

### learningphysics

Get the velocity of the helicopter by taking the derivative dh/dt... then plug in the time... and you get v1.

3. Sep 16, 2007

### Vii

Oooh, now why didn't I think of that before? Thank you!

But I've arrived at another problem: how to you tackle the height? Do you just use $$3.00t^3$$ to replace d, rearrange and get solve the quadratic equation?

i.e.

$$3.00t^3=v_1t+\frac{1}{2}at^2$$

4. Sep 16, 2007

### learningphysics

No that's not right.

The 3.00t^3 only works for the moment the object is released... after that it just falls downwards (it is no longer attached to the helicopter)...

Use the 3.00t^3 to get the initial conditions... ie the objects initial height and velocity... but after that you don't need it.

5. Sep 16, 2007

### Vii

Sorry. I'm afraid I still don't get it. I'm really horrible at Physics...

So, I get the initial conditions (the height and velocity) and substitute it into the equation and solve for t?

6. Sep 16, 2007

### learningphysics

Yes, just like in your initial post... everything is right except you assumed v1 was 0... find v1... then do the problem just like in your first post.

7. Sep 16, 2007

### Vii

Oh! Thank you so much!