What is the Time of Impact for a Falling Mailbag?

[ronaldh12]

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 x_f =(1/2)at² + v_ot + h_o

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.

 

[Doc Al]

What's the initial speed of the dropped mailbag? (How fast was the helicopter moving when it dropped the bag?)

 

[Hootenanny]

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 x_f =(1/2)at² + v_ot + h_o

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

Edit: Dammit Doc. :grumpy:

 

[ronaldh12]

I've given you all the information that I have. So, is there any way to find the velocity of the helicopter?

 

[Hootenanny]

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 ...

 

[LawrenceC]

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.

 

[Hootenanny]

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.

 

[ronaldh12]

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.

 

[LawrenceC]

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.

 

[Doc Al]

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.

 

[Doc Al]

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.)

 

[LawrenceC]

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.

 

[ronaldh12]

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.

 

[Hootenanny]

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 .

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 :grumpy:

 

[ronaldh12]

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. :]

 

[Clever-Name]

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.