From: rlamont at POSTOFFICE.PROVIDENCE.EDU (Bob LaMontagne)
Date: Fri Dec 31 14:43:00 2004
Ken - If h=0 the q = 1/2 arccos(1/infinity) = 1/2 arccos(0) = 1/2 90degrees
= 45.
Happy New Year to all -
Bob at PC
-----Original Message-----
From: Forum for Physics Educators [mailto:PHYS-L@list1.ucc.nau.edu] On
Behalf Of SSHS KPHOX
Sent: Friday, December 31, 2004 10:00 AM
To: PHYS-L@LISTS.NAU.EDU
Subject: Re: projectiles
Forum for Physics Educators <PHYS-L@list1.ucc.nau.edu> writes:
The angle that maximizes R is:
q = (1/2)*arccos(1/(1 + v^2/(g*h)))
Have I been on vacation too long? If h = 0 this should give q = 45°. But
at h=0, (v^2/g*h) is undefined. Am I way off in my thinking?
I think I remember once getting something like what David got (R =
(v^2/g)*cos(q)*sin(q)*(1 + sqrt(1 + 2*g*h/(v*sin(q))^2)))
and despairing that I could do the steps between to get a neat solution.
Today I would use the graphing calculator to plot the function and look
for zero slopes.
Back to work on Monday and I hope to have my head working right by then.