Re: Problem

[Bob Sciamanda <trebor@VELOCITY.NET>]

Hugh Haskell wrote:
> . . .
> 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.
Hugh, the square of the speed is given by

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