Free Falling Objects: Find Time for Mailbag to Reach Ground

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Homework Help Overview

The problem involves a helicopter releasing a mailbag, with the height of the helicopter described by a polynomial function of time. Participants are tasked with determining the time it takes for the mailbag to reach the ground after its release.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the initial height of the mailbag and the correct interpretation of the height function. There are attempts to apply kinematic equations to find the time of descent, with some questioning the initial conditions and the use of calculus for velocity.

Discussion Status

There is an ongoing exploration of the problem with various interpretations of the height function and its implications for the initial velocity. Some participants have offered corrections and adjustments to the calculations, while others are verifying the results without reaching a consensus.

Contextual Notes

Participants note the importance of including units in calculations and express concerns about potential point deductions for missing units. There is also a mention of the need to clarify the correct form of the height function, which has been amended during the discussion.

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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
Vi = 6.00t^2 at 2 = 24 m/s not sure about this

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

Did I go about this right? Please help.
 
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pberardi said:

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
Vi = 6.00t^2 at 2 = 24 m/s not sure about this

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

Did I go about this right? Please help.

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?
 
pberardi said:
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
?
 
pberardi said:
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
 
Last edited:

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