Experiment with forces - Formula question

AI Thread Summary
The discussion revolves around a physics experiment involving a ball of mass 530g, where a force of 7N was applied to determine the height from which it fell over 3 seconds. The initial formula used, h = (force acted * time) / mass, was found to be incorrect due to missing gravity and incorrect units. Participants clarified that the correct equation for height in free fall is h = (gt^2)/2, where g represents gravitational acceleration. The conversation also highlighted the confusion surrounding the application of forces and the need for proper understanding of kinematic equations. Overall, the experiment revealed the importance of using established physics principles to avoid errors in calculations.
-Physician
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Homework Statement


I took a ball of mass= 530g and someone held it in a high for me, i wanted to find that high and what i did is :
1. act a force of 7N on it
2. the time of ball going down was 3s
i wanted to find the high and what i did is:
##high=force acted * time / mass##
when i calculated that, the units were wrong and i squared the time. What i won was this:
##h=\frac{Ft^2}{m}## and i got
##m=\frac{\frac{kgm}{s^2} * s^2}{kg}##
(kg simplified with kilogram, the s^2 with s^2 , and i won m)
I did the same thing again and the high was exactly as calculated ( 120.7547169811321m)(121m)
But I think there's still something wrong in formula because there's no gravity in formula , but then my friend said ##a=F/m## but in this case we have a high so gravity would be F/m, and we would still win ##h=gt^2## . Is all this right? thanks for reading !
 
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You mean the ball was in a free fall from this height?

The simplest equation you can get is h = \frac{gt^2}{2}. Your equation h = \frac{Ft^2}{m} is exactly this (after simplification), and the acceleration of gravity is hidden in the force:

h = \frac{Ft^2}{m} = \frac{mgt^2}{m} = gt^2,

though 2 in the denominator is missing.

You can also try the conservation of energy: \frac{mv^2}{2} = mgh \rightarrow h = \frac{v^2}{2g} but then you would have to know the final velocity (before hitting the ground).
 
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So my formula is wrong?
 
-Physician said:
act a force of 7N on it
This was a vertically-directed force? What arrangement did you use to cause 7N to continuously act on the ball while it was falling? How did you neutralize gravity for your experiment?
the time of ball going down was 3s
i wanted to find the high and what i did is:
##high=force acted * time / mass##
So you invented your own equationhttps://www.physicsforums.com/images/icons/icon5.gif And you're surprised that it seems to give the right answer?
I'm surprised, too, but can't explain it. http://img140.imageshack.us/img140/7701/questionicon.gif

The equation you need is: s = ut + ½·at2
 
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As I see, you didn't read all the post bro, the high=force acted *time/mass was wrong, next time read all the post...
s=ut+at^2/2, it's v_0t not ut , or vt - at^2/2
 
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