Re: Problem

[Bob Sciamanda <trebor@VELOCITY.NET>]

----- Original Message -----
From: "Hugh Haskell" <hhaskell@MINDSPRING.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, September 18, 2001 4:21 PM
Subject: Re: Problem


> At 15:12 -0400 9/18/01, Bob Sciamanda wrote:
>
> > 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
>
> I think not, Bob. If you take the derivative of V(t) (the square root
> of your espression above), I think that you will find that it equals
> -g, which it should, since this is in a uniform grav. field, by the
> statement of the problem.

Hugh:
I get dV/dt =( -g/V )*(Vo*Sin TH -gt);  where V is the speed.
This goes through zero at t=(Vo*SinTH)/g.

The derivative of the speed is NOT the same as the magnitude of the
acceleration except, at times,  by coincidence!
> Furthermore, take the special case when TH=90 degrees, i.e., a
> vertical trajectory. Are you telling me that when the object gets to
> the very top it's acceleration goes to zero? IF so, how does it get
> back down?

Hugh,
The derivative of the speed is NOT the same as the magnitude of the
acceleration except, at times,  by coincidence!

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor