Velocity in Vacuum: Feather Dropped, 0.30 secs

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A feather dropped in a vacuum experiences an acceleration of 9.8 m/s² due to gravity. After 0.30 seconds, its velocity is calculated to be approximately 2.94 m/s. The distance fallen in that time is determined using the equation d = 1/2at², resulting in about 0.441 meters. The discussion also includes clarification on how to format superscripts and subscripts in forum posts. Overall, participants assist with calculations and formatting questions related to the physics of free fall.
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Ok:

Note: Acceleration due to gravity: g = 9.8 m/s

A feather is dropped inside a vacuum chamber.

A) What is the Velocity after 0.30 sec?

B) How far has it fallen after 0.30 sec?

I'm stuck on this.. it seems simple.. but maybe too simple. I think there is more too it. Any help?
 
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According the the policy one is supposed to show ones efforts before asking for assistance.
 
Umm.. I said I was stuck. I do not know where to begin at? Help me?
 
Acceleration is the rate of change of velocity with respect to time. Now, if an object is accelerating at 9.8 m/s/s (note the units) how fast will it be going after 1 second? 0.3 seconds?
 
after 1 sec it will go 9.8 right? and after .3 sec it will go 2.94.. i think.
 
Carnivean said:
after 1 sec it will go 9.8 right? and after .3 sec it will go 2.94.. i think.
Sounds good to me. Now, for the second question what kinematic equations do you know?
 
for distance I have d = 1/2at (the 't' is squared , but I don't know how to type a small '2' next to it.. if you could show me that too please) this is an equation for the problems where acceleration info is already is included.

where initial velocity is included: d= InitialVelocity x (t) + 1/2a x (t squared)
 
Carnivean said:
for distance I have d = 1/2at (the 't' is squared , but I don't know how to type a small '2' next to it.. if you could show me that too please) this is an equation for the problems where acceleration info is already is included.

where initial velocity is included: d= InitialVelocity x (t) + 1/2a x (t squared)
Your first equation looks applicable here :smile:. To type superscript text simply enclose the text in [#sup#] [#/sup#] tags (without the #).
 
ok I got .441 meters for the distance which I think sounds right. Almost a half a meter. But could you give me an example of how to do the superscript thing again? Is it [/then the number]? or [#the number#] ? Just show me an example of how you would type it.
 
  • #10
Carnivean said:
ok I got .441 meters for the distance which I think sounds right. Almost a half a meter. But could you give me an example of how to do the superscript thing again? Is it [/then the number]? or [#the number#] ? Just show me an example of how you would type it.

For example if you wanted to type x2, you would type:
Code:
x[#sup]2[/sup#]
Or for subscript;
Code:
x[#sub]2[/sub#]
Again, without the # marks.
 
  • #11
ok thanks soo much for your help.. you don't know how much it is appreciated. I will be back many a times probably until about january.

let me try - x 2
 
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