| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
25 Sep 2006 07:49:56 PM |
| Object: |
Please help me!!!! |
I am trying to solve a problem for my physics class and I am
consistently getting the damn wrong answer. Here is the problem:
A plane is flying horizontally with speed 204 m/s at
height 6720 m above the ground, when a package is thrown out at 2.2 m/s
downward. Neglecting air resistance and assuming acceleration due to
gravity is 9.8 m/s^2 what is the horizontal distance from release to
impact?
Answer in units of meters and assume that down is the negative y
direction.
I have attacked this several ways and the derivation I am consistently
getting is:
x=v*sqrt((2y)/g) or x=v*sqrt((-2y)/g) , depending on where I set
y=0
v=initial speed in the positive x direction
I dervied this equation using the kinematic equations. The homework
server is telling me that I am wrong, But I don't see any other way to
do it.
I can no longer get credit for the question so I am turning what I
believe to my own stupidity over to the forum for examination.
All comments, except for the extremely rude ones are welcome.
.
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| User: "Sorcerer" |
|
| Title: Re: Please help me!!!! |
25 Sep 2006 07:58:34 PM |
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<rehamkcirtap@gmail.com> wrote in message
news:1159231796.887542.187070@d34g2000cwd.googlegroups.com...
|I am trying to solve a problem for my physics class and I am
| consistently getting the damn wrong answer. Here is the problem:
|
| A plane is flying horizontally with speed 204 m/s at
| height 6720 m above the ground, when a package is thrown out at 2.2 m/s
| downward. Neglecting air resistance and assuming acceleration due to
| gravity is 9.8 m/s^2 what is the horizontal distance from release to
| impact?
| Answer in units of meters and assume that down is the negative y
| direction.
|
| I have attacked this several ways and the derivation I am consistently
| getting is:
|
| x=v*sqrt((2y)/g) or x=v*sqrt((-2y)/g) , depending on where I set
| y=0
|
| v=initial speed in the positive x direction
|
| I dervied this equation using the kinematic equations. The homework
| server is telling me that I am wrong, But I don't see any other way to
| do it.
|
| I can no longer get credit for the question so I am turning what I
| believe to my own stupidity over to the forum for examination.
|
| All comments, except for the extremely rude ones are welcome.
|
The vertical acceleration is used to find time.
Work out the time t to hit the ground and then the horizontal
distance is 204 m/s * t.
Androcles
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| User: "Ben Rudiak-Gould" |
|
| Title: Re: Please help me!!!! |
25 Sep 2006 08:12:57 PM |
|
|
wrote:
A plane is flying horizontally with speed 204 m/s at
height 6720 m above the ground, when a package is thrown out at 2.2 m/s
downward. Neglecting air resistance and assuming acceleration due to
gravity is 9.8 m/s^2 what is the horizontal distance from release to
impact?
I have attacked this several ways and the derivation I am consistently
getting is:
x=v*sqrt((2y)/g) or x=v*sqrt((-2y)/g), depending on where I set y=0
There are two speeds in the problem, but your answer only mentions one
speed. I think you forgot to consider the initial downward velocity of the
package.
-- Ben
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| User: "PD" |
|
| Title: Re: Please help me!!!! |
25 Sep 2006 08:51:41 PM |
|
|
wrote:
I am trying to solve a problem for my physics class and I am
consistently getting the damn wrong answer. Here is the problem:
A plane is flying horizontally with speed 204 m/s at
height 6720 m above the ground, when a package is thrown out at 2.2 m/s
downward. Neglecting air resistance and assuming acceleration due to
gravity is 9.8 m/s^2 what is the horizontal distance from release to
impact?
Answer in units of meters and assume that down is the negative y
direction.
I have attacked this several ways and the derivation I am consistently
getting is:
x=v*sqrt((2y)/g) or x=v*sqrt((-2y)/g) , depending on where I set
y=0
v=initial speed in the positive x direction
This formula you've derived only works if the package is released,
dropped, let go. It's not. It's hurled downward. When the package
leaves the plane, it is going 204 m/s horizontally and 2.2 m/s
vertically.
I dervied this equation using the kinematic equations. The homework
server is telling me that I am wrong, But I don't see any other way to
do it.
I can no longer get credit for the question so I am turning what I
believe to my own stupidity over to the forum for examination.
All comments, except for the extremely rude ones are welcome.
.
|
|
|
|
| User: "" |
|
| Title: Re: Please help me!!!! |
25 Sep 2006 09:46:54 PM |
|
|
In article <1159231796.887542.187070@d34g2000cwd.googlegroups.com>, "rehamkcirtap@gmail.com" <rehamkcirtap@gmail.com> writes:
I am trying to solve a problem for my physics class and I am
consistently getting the damn wrong answer. Here is the problem:
A plane is flying horizontally with speed 204 m/s at
height 6720 m above the ground, when a package is thrown out at 2.2 m/s
downward. Neglecting air resistance and assuming acceleration due to
gravity is 9.8 m/s^2 what is the horizontal distance from release to
impact?
Answer in units of meters and assume that down is the negative y
direction.
I have attacked this several ways and the derivation I am consistently
getting is:
x=v*sqrt((2y)/g) or x=v*sqrt((-2y)/g) , depending on where I set
y=0
v=initial speed in the positive x direction
I dervied this equation using the kinematic equations. The homework
server is telling me that I am wrong, But I don't see any other way to
do it.
I can no longer get credit for the question so I am turning what I
believe to my own stupidity over to the forum for examination.
All comments, except for the extremely rude ones are welcome.
Well, you have done enough to warrant a pointer. You forgot the
initial velocity, in the y direction. Setting (just to have some agreed
upon coordinates) y=0 on the ground with the positive y direction
upwards, the general eqaution for y is
y = y(0) + v_y(0)*t - g*t^2/2
In your case y(0) is the initial elevation, 6720 m, and v_y(0) is the
initial velocity in the y direction, given above as -2.2 m/s. The
time t it takes to hit the ground is obtained from solving the above
for y = 0 (do you understand why?). Once you found the time, then you
can find the horizontal distance traveled.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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