# The instantaneous velocity of a freely falling object

 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?
 HW Helper P: 5,341 Welcome to PF. You can approach average velocity either way. V = a*t => Vavg = 1/2*a*t or X = 1/2*a*t2 Vavg = Xtot/Ttot = 1/2*a*t2/t = 1/2*a*t V2 = 2*a*x = 2*a*(1/2*a*t2) = a2t2 V = a*t They are all interrelated.
 P: 4 The answers came up slightly different would it matter?
HW Helper
P: 5,341
The instantaneous velocity of a freely falling object

 Quote by sammyj The answers came up slightly different would it matter?
There is no difference that I see. Acceleration is due to gravity in both cases.

The bottom equation uses g = 10 instead of 9.8.

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