Max Rage of a Projectile Uphill

[Tom McCurdy]

I am working on some physics on a site I found and was wondering if I could tap into some of the wisdom that seems embeded in the forums.

The problem I am working on is problem three from
http://www.math.rutgers.edu/~costin/291/w3.pdf

All I am interested in doing is show how to optimize the uphill range.

I have been doing some inital attacking but don't seem to be getting anywhere. I know the answer from that site tan 2 \alpha = -cot \theta

Here is some of the stuff that I have been starting out doing
y/x=tan\theta
y= x tan \theta
R/x= sec \theta
R = (Vot+1/2at^2) sec \theta
\frac {v^2-V_o^2}{{2*9.8}}=x

I am not very sure on how many of these are true because I haven't had time fully to reexamine them and I was trying to force things together that probably need to be placed. I will continue to work on this problem as I just started but If anyone can sucessfully show me how to come up with

[Pyrrhus]

Try to make a formula for the distance uphill, then apply calculus to get the maximiun value for the alpha angle. Consider intersection....

xtan\theta = tan\alpha x - \frac{g}{2v_{o}^{2}cos^{2}\alpha}x^{2}


0 = (tan\alpha - tan\theta)x - \frac{g}{2v_{o}^{2}cos^{2}\alpha}x^{2}


0 = x((tan\alpha - tan\theta) - \frac{g}{2v_{o}^{2}cos^{2}\alpha}x)

so one x=0 and the other, you know....

after finding both X and Y of intersection, a simple apply of the distance formula with 0,0 and X,Y can get you d, after that you could find the max value for alpha using differential calculus.

[Tom McCurdy]

Still a little lost... I am will look to see if there is any other help when I wake up...;
for example I am just taking my first calc course Calc BC how would us use calc to ge the max value for the alpha angle...



ty for your response Cyclovenom

[Pyrrhus]

By the way, i edited with some of the work.

[Tom McCurdy]

There must be a way to do it with out differential calculus... the physics course I am taking is made to have no math beyond deravatives and integration in calculus. I was wondering if someone could show me how to derrive

tan 2 \alpha = -cot \theta

from the example

[HallsofIvy]

First, ignore the slope. A projectile (even a "raging" projectile!) fired with initial speed v0 and angle to the horizontal α has trajectory given by
x= v0cosα t and y= -(g/2)t2+ v0sinα t.

If the ground were level, "hitting the ground" would correspond to y= 0

Here, the "ground" is given by y= x tanθ so the projectile will hit the ground when tan&theta(v0cosα t)= -(g/2)t2+ v0sinα t

You could solve that quadratic equation for t in terms of α, which is fixed, and θ and then find the θ that maximizes x. I'n sure HOW you would do that without using the derivative!

[Tom McCurdy]

Method One:
In this method please note that in the picture of the problem
\alpha is replaced by \beta
\theta is replaced by \alpha

Work

y=xtan\alpha
x=Vocos\beta t
y=Vosin\beta=-1/2gt^2

Vosin\Beta t = 1/2gt^2=xtan\alpha=Vocos/beta t tan \alpha
t^2-\frac2{2}{{g}}(Vosin\beta-Vocos\beta tan\alpha)=0
t = (t-\frac2{2}{{g}}(Vosin\beta-Vocos\beta tan\alpha))=0
t=0 or t= \frac{2Vo}{{g}}(sin\beta-cos\betatan\alpha
sec\alpha=\frac{R}{{x}}
R=xsec\alpha
=Vocos\beta t sec\alpha
= Vocos\beta sec\alpha t
=Vocos\beta sec\alpha (\frac{2vo}{{g}}(sin\beta-cos\beta tan\alpha))
=\frac{2vo^2}{{g}}\frac{sin\beta cos\beta}{{cos\alpha}}-\frac{cos^2\beta}{{cos\alpha}}\frac{sin\alpha}{{co s\alpha}}
\frac{2Vo^2}{{g}} (\frac{sin\beta cos\beta cos\alpha - cos^2\beta sin\alpha}{{cos^2\alpha}}
\frac{2Vo^2}{{g}}(\frac{sin\beta cos\alpha - cos\beta sin \alpha }{{cos^2\alpha}})cos\beta
{2Vo^2}{{g}} \frac{{sin(alpha-\beta)cos\beta}{{cos^2\alpha}}
R= \frac{2Vo^2}{{g}}(\frac{sin(\alpha - \beta)cos\beta}{{cos^2\alpha}})
R= \frac{2Vo^2}{{g}}(\frac{sin(2\beta-\alpha)-sin\alpha}{{cos^2\alpha}}
since everything is constant minus sin(2/beta-/alpha) you must maximize it[/tex]

2\beta = \frac{\pi}{{2}}+\alpha
\beta = \frac{\pi}{{4}}+\frac{\alpha}{{2}}

[Tom McCurdy]

Does this seem alright

[Pyrrhus]

Tom, will you please fix the Latex? oh and Yes the answer is correct.

[Tom McCurdy]

I can't edit the post any more otherwise I would