Velocity in Vacuum: Feather Dropped, 0.30 secs

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Homework Help Overview

The original poster presents a problem involving the motion of a feather dropped in a vacuum, specifically asking for its velocity and distance fallen after 0.30 seconds, given the acceleration due to gravity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the basic principles of acceleration and velocity, with some attempting to calculate the velocity after 0.3 seconds. Others introduce kinematic equations for calculating distance fallen, while some express uncertainty about how to format mathematical expressions.

Discussion Status

Participants are actively engaging with the problem, exploring different aspects of motion under gravity. Some have provided calculations and equations, while others seek clarification on formatting and expressing mathematical terms. There is no explicit consensus on the final answers yet.

Contextual Notes

The original poster indicates feeling stuck and questions whether the problem may be more complex than it appears. There is a mention of forum policy regarding showing efforts before seeking help.

Carnivean
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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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Welcome to the Forums,

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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