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The instantaneous velocity of a freely falling object

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sammyj
#1
Nov13-08, 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 = gt squared
* 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

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LowlyPion
#2
Nov13-08, 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*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.
sammyj
#3
Nov13-08, 10:37 AM
P: 4
The answers came up slightly different would it matter?

LowlyPion
#4
Nov13-08, 11:35 AM
HW Helper
P: 5,343
The instantaneous velocity of a freely falling object

Quote Quote by sammyj View Post
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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