Acceleration, Velocity, and Position

we all know that the velocity of an object thrown vertically upwards at highest point is zero and acc. is 9.8m/s^2 downwards. Also we know instantaneous acc. =V.dV/dX now putting the value V=0 in the formula we get acc=0 where V=0 i.e. at highest point. Now my question is what is the fact behind this double answer?? Is acc is 9.8m/s^2 or zero??? plzz help me in this ques...
- Kushmeet singh saluja (age 17)
Agra,india

Good question. Look at that dV/dX term. Remember that dV/dT is constant, so as V=dX/dt gets small (approaching the top) dV/dX must get big. Right at the top your formula
VdV/dX becomes zero*infinity, and thus doesn't have a defined value. At that point, only the more basic fixed acceleration can give you the answer.

Mike W.

(published on 02/14/2013)

Thank you Mike. You said the the formula V.dV/dX doesn't give the the defined value in this condition this means the formula fails in this condition. but so far i have studied in the books that this formula l satisfy at all the points. I have nt studied any exceptions. so formula should satisfy the conditions.
- Kushmeet singh saluja (age 17)
Agra,India

Kushmeet-

If those books say that the formula works in every case, they are wrong. We've just demonstrated an exception.

Mike W.

(published on 02/16/2013)