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[Phys-L] Re: projectiles



Here's a simple formula and a rapidly convergent algorithm
for your case. You can work it out easily for yourself if you just
remember that horizontal distance traveled is a fine proxy for the elapsed
time:
1 Define a parameter q=2gH/V^{2} (V^{2} is the TeX way to write the
exponent 2; V is the magnitude of the intitial velocity vector)
2. The launch angle a for maximum range is given by:
Cos(2a)= qSin(a)/[Sin(a)+A)
where A^{2} = Sin^{2}(a) +q
3. For the algorithm, put a starting guess for a on the right hand side.
The value you get for Cos(2a) gives you an improved value for a. 45
degrees is a good starting guess. This procedure converges to three
places in one iteration for q=.4. The angle for max range in that case is
40.2 degrees.
There may be a condition on the starting angle for values of q>3.
The condition may be that Sin(a)<1/sqrt(q-2). I'll leave it for others to
explore this condition.
Regards,
Jack

On Thu, 30 Dec 2004, Anthony Lapinski wrote:

Neglecting air resistance, we all know that the maximum range for a
projectile occurs when the launch angle is 45° on level ground. But what
if the object is launched from a cliff of height H above the ground? Here,
the maximum range occurs at less than 45°. I've searched most of the
college texts I have, but I can't find a "range formula" that has the
height included. Is there such an equation with an initial height (similar
to R = v^2 sin2q/g on level ground)?




--
"Trust me. I have a lot of experience at this."
General Custer's unremembered message to his men,
just before leading them into the Little Big Horn Valley