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.
One second is the time that elapses during 9,192,631,770 (9.192631770 x 10^9) cycles of the radiation produced by the transition between two levels of the cesium 133 atom.
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 =...