
#1
Mar812, 12:59 PM

P: 85

1. The problem statement, all variables and given/known data
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 ! 



#2
Mar812, 02:53 PM

P: 19

You mean the ball was in a free fall from this height?
The simplest equation you can get is [itex]h = \frac{gt^2}{2}[/itex]. Your equation [itex]h = \frac{Ft^2}{m}[/itex] is exactly this (after simplification), and the acceleration of gravity is hidden in the force: [itex]h = \frac{Ft^2}{m} = \frac{mgt^2}{m} = gt^2[/itex], though [itex] 2[/itex] in the denominator is missing. You can also try the conservation of energy: [itex] \frac{mv^2}{2} = mgh \rightarrow h = \frac{v^2}{2g}[/itex] but then you would have to know the final velocity (before hitting the ground). 



#3
Mar912, 12:08 PM

P: 85

So my formula is wrong?




#4
Mar912, 08:29 PM

HW Helper
P: 4,715

Experiment with forces  Formula questionI'm surprised, too, but can't explain it. The equation you need is: s = ut + ½·at^{2} 



#5
Mar1112, 06:52 AM

P: 85

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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