The hot-air ballon intersection question.

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Homework Help Overview

The problem involves a hot-air balloon rising at a constant rate and a camera being tossed upward. The goal is to determine the height of the balloon when the camera reaches it, given specific initial velocities and heights.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up equations for the heights of both the balloon and the camera to find the time when they are equal. There is a discussion about the significance of rounding and significant figures in the final answer.

Discussion Status

Participants are exploring the calculations and discussing the implications of significant figures. Some guidance has been offered regarding the rounding of the answer, but there is no consensus on the correct final answer due to differing interpretations of significant figures.

Contextual Notes

The problem is constrained by the requirement to express the answer using two significant figures, which has led to confusion among participants regarding the correct formatting of the answer.

aeabdo11
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I am close to positive I am going at this problem the correct way, but there seems to be some error somewhere. This problem is from online homework.

Homework Statement



A hot-air balloon has just lifted off and is rising at the constant rate of 2.2 m/s. Suddenly one of the passengers realizes she has left her camera on the ground. A friend picks it up and tosses it straight upward with an initial speed of 12 m/s.

If the passenger is 2.5 m above her friend when the camera is tossed, how high is she when the camera reaches her?
Express your answer using two significant figures.

Homework Equations



X=Xo+volt+(.5)at^2

The Attempt at a Solution



So I know I need to set the two equations equal and solve for time.

2.5+2.2t=12t-(.5)(9.81)t^2

Solving for t (using quad. equation) gives solutions as t=.3 , t= 1.7

Now I plug the time into the first equation and solve for x for the hot-air balloon at .3 seconds.

2.5+2.2(.3) = 3.16 m

As should be the case, I get the same answer when figuring x for the camera:

12(.3)-(.5)(9.81)(.3)^2= 3.158 m

When I plug this answer into the online homework it says, "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures."

I have tried the following answers, none of which work: 3.1, 3.15, 3.2

PLEASE HELP ME FIND WHAT I'M MISSING PLEASE!

Thanks a lot.
-Anthony
 
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Your equation is correct. Since the online homework website is saying that you're having some kind of rounding error, I went on checking the values. It appears that the actual value should be approximately 3.16045. Try to input 3.16 as an answer there.
 
I would have already tried that answer but the question asks for the answer to be in only two significant figures. "3.16" is three figures is it not (this significant figure thing is a bit confusing)?
 
aeabdo11 said:
I would have already tried that answer but the question asks for the answer to be in only two significant figures. "3.16" is three figures is it not (this significant figure thing is a bit confusing)?
In case of 'two significant figures', the answer should definitely be 3.2
 
Yeah Iv'e tried it...says incorrect.
 

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