[Phys-L] Re: projectiles

[cblreno at ADELPHIA.NET (Charles Reno)]

Using TI-82, 83, 83+ or 84(in degree mode):
First, with 40 -> V, 9.8 -> G.
Method I
Use Y1 =3D  fMax(Vcos(T)(Vsin(T)+sqrt((Vsin(T))^2+2GX))/G,T,0,45)
X is the height and the Y1 function gives the critical angle.
On the home screen. Y1(0) will use 0 as the height and figure out 45*
Check Y1(0) is 45*,   Y1(100) is 33.8372137* and Y1(240 is 26.7382137=
*
Now, turn Y1 off and define Y2, in terms of height,X, to be the max r=
ange:
Y2 =3D Vcos(Y1(X))(Vsin(Y1(X))+sqrt((Vsin(Y1(X)))^2+2GX))/G
On the home screen,Y2(0) will use 0 as the height and figure the max =
range
   to be 163.249
Y2(100) is 243.51561, Y2(240) is 324.0724
Using window such as   0 <=3D X <=3D 470,   0 <=3D Y <=3D 470
and Xres =3D 4 because I'm impatient, you can examine the graph of be=
st range=20
vs height.

Method II
It has been shown that for any height, the best range will be when th=
e=20
landing angle
    is the complement of the launch angle.
Using conservation of mechanical energy and constancy of horizontal v=
elocity=20
component
leads on to solve the equation:
(V^2+2GH)Tan(theta)^2 =3D V^2 for critical angle theta.
Use the solver( math 0 on the keyboard )
Use the three cases solved/shown above( H=3D0, 100, 240  )
      to verify the same critical angle.

I agree, this problem is a nightmare with classical techniques.
But, the tools of the handheld technology give a ready solution.

Enjoy,
Charlie
ps I have done and will do teacher training workshops with graphing c=
alc and CBL.
I have done the Connecting Math and Science week many times.

----- Original Message -----=20
=46rom: "Anthony Lapinski" <anthony_lapinski@PDS.ORG>
To: <PHYS-L@LISTS.NAU.EDU>
Sent: Thursday, December 30, 2004 11:44 AM
Subject: projectiles


 I've searched most of the
> college texts I have, but I can't find a "range formula" that has t=
he
> height included. Is there such an equation with an initial height (=
similar
> to R =3D v^2 sin2q/g on level ground)?