Which Distance is Correct for a Free Falling Object?

[arunkumarg]

I have a simple question may sound stupid for most of you:

For a free falling object v=9.8 m/sec. which of the distance traveled per sec is correct:

Formula 1:

d = vt
d = 9.8/1 = 9.8 m

Formula 2:
d = 1/2 a*t^2

acc of a free falling object is 9.8m/sec^2

d = 1/2 9.8*1^2
d = 4.9 m

This is a bit confusing.

 

[Doc Al]

For a free falling object v=9.8 m/sec.

No, the acceleration of a free falling object is 9.8 m/s^2. The velocity is not constant.

Formula 1:

d = vt
d = 9.8/1 = 9.8 m

This is incorrect as it assumes a constant velocity.

Formula 2:
d = 1/2 a*t^2

acc of a free falling object is 9.8m/sec^2

d = 1/2 9.8*1^2
d = 4.9 m

This is correct. That's the distance fallen in the first second.

 

[arunkumarg]

Thanks for your reply.

When I say V = 9.8m/sec. its the instantaneous velocity (for the 1st Sec).

Hence, for the 1st sec the dist must be 9.8m.

I know I am missing the point somewhere.

 

[Doc Al]

When I say V = 9.8m/sec. its the instantaneous velocity (for the 1st Sec).

That's the instantaneous velocity at the end of the 1st second. It starts out with zero velocity at t = 0 and ends up going 9.8 m/s at t = 1 sec.

Hence, for the 1st sec the dist must be 9.8m.

That would be true if the speed were constant for the entire second. But it's not. Since it starts from rest, the average speed during the 1st second is only 4.9 m/s.

 

[arunkumarg]

Sounds good for me... Thanks pal