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Acceleration due to gravity question 
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#1
Aug714, 04:21 AM

P: 7

Hi guys
To me I sound stupid, but I'm probably just overthinking the question. I know acceleration due to gravity is 9.8m/s/s. But if the distance is less than a metre or takes less than a second, wouldn't the acceleration be less than 9.8? If so, how is that worked out? If not, then is it correct to say that if something falls for a distance of 0.7m and comes to a stop, the acceleration is still 9.8m/s/s? 


#2
Aug714, 04:34 AM

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#3
Aug714, 05:14 AM

P: 7

Ok thanks. Next question. If the object bounces back up, is the acceleration still the same? And how do I go about working out the time it takes after impact to bounce to a height? The only details I've got is what I used to calculate velocity and momentum at impact.



#4
Aug714, 05:51 AM

Thanks
P: 5,869

Acceleration due to gravity question
Acceleration is not defined in terms of distance and time. It is defined in terms of velocity and time. To determine average acceleration, you take initial velocity and final velocity, and divide it by time.
You can relate acceleration and distance, too, but then you need more details. For example, if you can assume that your acceleration is constant, as in the case of acceleration due to gravity close to the Earth, then you can use the SUVAT equations. 


#5
Aug714, 05:52 AM

P: 7

[tex]E = \frac{1}{2} m v^2+m g h[/tex] And using the height at impact (h=0) to calculate the kinetic energy of the system. At height = h you can calculate the velocity and using [itex]v_h = v_0  gt[/itex] you can calculate the time. 


#6
Aug714, 12:39 PM

P: 18




#7
Aug714, 12:50 PM

Math
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Thanks
PF Gold
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#8
Aug714, 01:57 PM

P: 1,002

http://en.wikipedia.org/wiki/Gravity_of_Earth 


#9
Aug714, 02:41 PM

P: 18

Check. Thanks for the additional information, both of you.



#10
Aug714, 05:08 PM

P: 7

Ohhhh I think I know why I was confused. I was thinking about changes in velocity. So the velocity will change even though acceleration is still the same, right?



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