Re: Problem

[Bob Sciamanda <trebor@VELOCITY.NET>]

To amplify a bit, qualitative (at most semi-quantitative) analysis alone
shows:
Since the horizontal velocity is constant and the vertical velocity turns
around after going through zero at the trajectory top - The speed begins
at a maximum (at firing), reduces in value to a minimum (at the trajectory
top), and then increases until impact.  This means that the speed vs time
function has a minimum, and so a zero derivative.  This occurs when the
projectile is at the trajectory top and the velocity is horizontal.

Bob
----- Original Message -----
From: "Bob Sciamanda" <trebor@VELOCITY.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, September 18, 2001 3:12 PM
Subject: Re: Problem


> 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