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