# Free falling objects

• pberardi

## Homework Statement

The height of a helicopter above the ground is given by h = 3.00t^2, 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

## Homework Equations

Xf = Xi + VxT + .5at^2

## The Attempt at a Solution

h = 24
a = -9.8

0 = 24 + 24t + .5(-9.8)t^2
t = 5.7

## Homework Statement

The height of a helicopter above the ground is given by h = 3.00t^2, 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

## Homework Equations

Xf = Xi + VxT + .5at^2

## The Attempt at a Solution

h = 24
a = -9.8

0 = 24 + 24t + .5(-9.8)t^2
t = 5.7

Wouldn't the initial height be 12? 2^2*3 = 12?

Also, do you do physics with calculus? If so, you could take the derivative of the height function to get the velocity function, and simply plug in for t. If not, you can only get the average velocity, which won't do you any good in this problem, since the helicopter is accelerating.

I am sorry, my mistake. I did do the calculus. But I wrote the height function incorrectly. It is supposed to be h = 3t^3

So based on this new information, is it correct?

I am sorry, my mistake. I did do the calculus. But I wrote the height function incorrectly. It is supposed to be h = 3t^3

So based on this new information, is it correct?
The initial height is correct, but the derivative of 3t^3 is 9t^2. With that in mind, the initial velocity would be 36.

Should be
0 = 24 + 36t + .5(-9.8)t^2
t = 7.96
?

Should be
0 = 24 + 36t + .5(-9.8)t^2
t = 7.96
?
That looks good.

Thank you sir.

They might deduct a couple points if units aren't shown in every step ...
0 = 24.0 m + (36.0 m/s) t + (1/2) (-9.8 m/s^2) t^2

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