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The instantaneous velocity of a freely falling object 
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#1
Nov1308, 09:28 AM

P: 4

What is the instantaneous velocity of a freely falling object 10 s after it is released from a position of rest? What is its average velocity during this 10 s interval? How far will it fall during this time?
I had asked a few friends for help and I keep getting mixed answers. v = gt 98m/s = 9.8m/s * 10 Average V = initial v + final v 2 Average V = 0m/s + 98m/s 2 Average V=49m/s Distance Traveled d = ½ gt squared ½ * 9.8m/s * 10 squared = 490m or Vf=Vo+at = 0msec+10msec2(10sec)= 100msec AverageVelocity=Vf+Vo2=100ms+0ms2=50msec X=Vot +12at2=0ms(10sec)+12(10msec2)(10sec)2=500 meters Who's on the right path? 


#2
Nov1308, 10:23 AM

HW Helper
P: 5,343

Welcome to PF.
You can approach average velocity either way. V = a*t => Vavg = 1/2*a*t or X = 1/2*a*t^{2} Vavg = Xtot/Ttot = 1/2*a*t^{2}/t = 1/2*a*t V^{2} = 2*a*x = 2*a*(1/2*a*t^{2}) = a^{2}t^{2} V = a*t They are all interrelated. 


#3
Nov1308, 10:37 AM

P: 4

The answers came up slightly different would it matter?



#4
Nov1308, 11:35 AM

HW Helper
P: 5,343

The instantaneous velocity of a freely falling object
The bottom equation uses g = 10 instead of 9.8. 


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