# The instantaneous velocity of a freely falling object

1. Nov 13, 2008

### sammyj

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?

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. Nov 13, 2008

### LowlyPion

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.

3. Nov 13, 2008

### sammyj

The answers came up slightly different would it matter?

4. Nov 13, 2008

### LowlyPion

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.