[Phys-L] Re: projectiles

[David_Bowman at GEORGETOWNCOLLEGE.EDU (David Bowman)]

Regarding Ken's question:

> 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°.

It does (assuming you know how to take limits).

> But at h=0, (v^2/g*h) is undefined. Am I way off in my thinking?

It is true that v^2/(g*h) diverges as h --> 0, but the relevant
quantity is 1/(1 + v^2/(g*h)) = g*h/(g*h + v^2) --> 0 as h --> 0 and
(1/2)*arcos(0) = [pi]/2 rad = 45 deg.

> 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.

When I extremized the above expression over q I first wrote it in
terms of the quantity s == 1 - cos(2*q), used a couple of double
angle formulae, and then minimized w.r.t. s which nicely eliminated
all the trig functions in the intermediate steps. After finding the
value of s that extremized R I wrote q = (1/2)*arcos(1 - s) to get
the corresponding value of q.  It also helped to write the problem
in terms of the dimensionless ratio p of the initial potential
energy to the initial kinetic energy, i.e. p = 2*g*h/v^2.

> Today I would use the graphing calculator to plot the function and
> look for zero slopes.

A picture/graph may be worth a thousand words/numbers, but a
formula is worth *many more* than a thousand picture/graphs.

> Back to work on Monday and I hope to have my head working right by
> then.

Moderation is a virtue.

> Happy New Year all.

Cheers.

> Ken Fox

David Bowman