What is the Time of Impact for a Falling Mailbag?

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Homework Help Overview

The problem involves determining the time it takes for a mailbag to reach the ground after being released from a helicopter. The height of the helicopter is described by a cubic function of time, and the question arises after the helicopter has been in motion for a specific duration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial height of the mailbag and the initial velocity at the moment of release. There are attempts to apply kinematic equations to find the time of impact, with some participants questioning the accuracy of their calculations and the interpretation of velocity.

Discussion Status

Some participants have provided guidance on calculating the instantaneous velocity of the helicopter at the time of release, suggesting the use of calculus. There is an ongoing exploration of the problem's wording and the assumptions made about the helicopter's motion.

Contextual Notes

Participants note that the problem may have been loosely stated, leading to confusion about the helicopter's motion and the necessary calculations. There is a recognition of the need for clarity regarding the instantaneous versus average velocity.

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Homework Statement



The height of a helicopter above the ground is given by h = 2.90t3, where h is in meters and t is in seconds. At t = 1.75 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Homework Equations



I used xf =(1/2)at2 + vot + ho

The Attempt at a Solution



To get the initial height, I used the t = 1.75 s in the h = 2.90t3 resulting in an original height of 15.5422 or 15.5. So, therefore I got: 0 = (1/2)(-9.8)t2 + (0)(t) + 15.5. Resulting in t = 1.78 seconds, which was incorrect and apparently off by more than 10 percent according to WebAssign.

Thank you for all your help.
 
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What's the initial speed of the dropped mailbag? (How fast was the helicopter moving when it dropped the bag?)
 
ronaldh12 said:

Homework Statement



The height of a helicopter above the ground is given by h = 2.90t3, where h is in meters and t is in seconds. At t = 1.75 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?

Homework Equations



I used xf =(1/2)at2 + vot + ho

The Attempt at a Solution



To get the initial height, I used the t = 1.75 s in the h = 2.90t3 resulting in an original height of 15.5422 or 15.5. So, therefore I got: 0 = (1/2)(-9.8)t2 + (0)(t) + 15.5. Resulting in t = 1.78 seconds, which was incorrect and apparently off by more than 10 percent according to WebAssign.

Thank you for all your help.
The initial velocity of the mailbag is non-zero :wink:

Edit: Dammit Doc.
 
I've given you all the information that I have. So, is there any way to find the velocity of the helicopter?
 
ronaldh12 said:
I've given you all the information that I have. So, is there any way to find the velocity of the helicopter?
Sure! You know the height [position] of the helicopter at time t ...
 
Because the altitude of the helicopter is given as a function of time, you could take the first derivative to get its upward velocity. Then solve the problem with that initial upward velocity. See if that gets you the answer you seek. Problem was loosely stated in my opinion.
 
LawrenceC said:
Because the altitude of the helicopter is given as a function of time, you could take the first derivative to get its upward velocity. Then solve the problem with that initial upward velocity. See if that gets you the answer you seek. Problem was loosely stated in my opinion.
I fail to see how the problem was loosely stated. All the required information was given, there is no ambiguity in the wording or in the physical system.
 
Hootenanny said:
Sure! You know the height [position] of the helicopter at time t ...

Okay, I calculated a velocity of 8.88 m/s, because the height is 15.5 m at t = 1.75.

(15.5-0)/1.75 to get the 8.88 m/s

So, I get

0 = (1/2)(-9.8)t2 + 8.88t +15.5

Solve for t with a polynomial function, and get t = -1.09 and 2.90. Time can't be negative and 2.90 isn't the right answer.

I feel like I'm clinically insane for not being able to solve this problem...lol.
 
Hootenanny said:
I fail to see how the problem was loosely stated. All the required information was given, there is no ambiguity in the wording or in the physical system.

Sounds like you might be the author of the problem... ; ). I would have stated it as 'the altitude of an ascending helicopter...' I feel that would have made it more clear to the student.
 
  • #10
ronaldh12 said:
Okay, I calculated a velocity of 8.88 m/s, because the height is 15.5 m at t = 1.75.

(15.5-0)/1.75 to get the 8.88 m/s
You've calculated the average speed of the helicopter as it rose to the given height. But what you need is the instantaneous speed at the moment the mailbag is released.

Hint: Calculus.
 
  • #11
LawrenceC said:
I would have stated it as 'the altitude of an ascending helicopter...' I feel that would have made it more clear to the student.
They give the height as a function of time, which is all the information needed. (Part of the problem is to be able to interpret what that means.)
 
  • #12
ronaldh12 said:
Okay, I calculated a velocity of 8.88 m/s, because the height is 15.5 m at t = 1.75.

(15.5-0)/1.75 to get the 8.88 m/s

So, I get

0 = (1/2)(-9.8)t2 + 8.88t +15.5

Solve for t with a polynomial function, and get t = -1.09 and 2.90. Time can't be negative and 2.90 isn't the right answer.

I feel like I'm clinically insane for not being able to solve this problem...lol.

You're not clinically insane. Your velocity is incorrect. The position is given as a function of time. All you've done is computed the average velocity over 1.75 seconds. You need the instantaneous velocity at 1.75 seconds. Think about it some more and think of calculus while you're doing your thinking.
 
  • #13
LawrenceC said:
You're not clinically insane. Your velocity is incorrect. The position is given as a function of time. All you've done is computed the average velocity over 1.75 seconds. You need the instantaneous velocity at 1.75 seconds. Think about it some more and think of calculus while you're doing your thinking.

Is it really that simple? And you want the instantaneous velocity, because the average velocity incorporates the time when there was no velocity. You want it at that time.

Makes sense. I appreciate all of your help.
 
  • #14
LawrenceC said:
Sounds like you might be the author of the problem... ; ). I would have stated it as 'the altitude of an ascending helicopter...' I feel that would have made it more clear to the student.
I'm just grumpy :smile:.

I'm the first to admit that there are some very poorly worded, misleading or frankly incorrect problems set by tutors (I've marked my fair share of insolvable problems). However, I do get a little ratty when students (I know you're not the student) complain about problems that are adequately worded and can be solved in a relatively straightforward manner. Admittedly, the acceleration of the helicopter was obfuscated, but perhaps that was the intention of the author of the problem.

As I said, I'm grumpy grad student who doesn't like marking :wink:
 
Last edited:
  • #15
Thank you all again. You're help is appreciated. Hopefully I won't have to come on here too often, but if I do you guys are the first place to help me deal with my problems. Physics related problems of course, but problems nonetheless. :]
 
  • #16
Have you made any further progress on the question? I'm not positive you are grasping the idea of instantaneous velocity properly at this point.
 

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