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Hugh Haskell wrote:
. . .Hugh, the square of the speed is given by
In the trajectory case, there are
terms involving *gt* in the expression for the speed, Hence the speed
is *not* constant, even if the component subject tot he acceleration
happens at some instant to be zero. The time derivative of the speed
is *not* zero, hence it *is* changing.
V(t)^2 = (Vo*CosTH)^2 +(Vo*SinTH-gt)^2
The time derivative of this goes through zero at the top of the
trajectory,
ie: at t = (Vo*SinTH)/g
Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor