# Homework Help: This question blows my mind

1. Aug 29, 2011

### ronaldh12

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

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?

2. Relevant equations

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

3. 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.

2. Aug 29, 2011

### Staff: Mentor

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

3. Aug 29, 2011

### Hootenanny

Staff Emeritus
The initial velocity of the mailbag is non-zero

Edit: Dammit Doc. :grumpy:

4. Aug 29, 2011

### ronaldh12

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

5. Aug 29, 2011

### Hootenanny

Staff Emeritus
Sure! You know the height [position] of the helicopter at time t ...

6. Aug 29, 2011

### 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.

7. Aug 29, 2011

### Hootenanny

Staff Emeritus
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.

8. Aug 29, 2011

### ronaldh12

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.

9. Aug 29, 2011

### LawrenceC

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. Aug 29, 2011

### Staff: Mentor

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. Aug 29, 2011

### Staff: Mentor

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. Aug 29, 2011

### LawrenceC

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. Aug 29, 2011

### ronaldh12

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. Aug 29, 2011

### Hootenanny

Staff Emeritus
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:

Last edited: Aug 29, 2011
15. Aug 29, 2011

### 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. :]

16. Aug 29, 2011

### 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.