# Max Rage of a Projectile Uphill

1. Sep 30, 2004

### Tom McCurdy

Help: Max Rage of a Projectile Uphill

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 [Broken]

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

Last edited by a moderator: May 1, 2017
2. Sep 30, 2004

### 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.

Last edited: Sep 30, 2004
3. Sep 30, 2004

### 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...

4. Sep 30, 2004

### Pyrrhus

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

5. Oct 1, 2004

### 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

6. Oct 1, 2004

### HallsofIvy

Staff Emeritus
First, ignore the slope. A projectile (even a "raging" projectile!) fired with initial speed v0 and angle to the horizontal &alpha; has trajectory given by
x= v0cos&alpha; t and y= -(g/2)t2+ v0sin&alpha; t.

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

Here, the "ground" is given by y= x tan&theta; so the projectile will hit the ground when tan&theta(v0cos&alpha; t)= -(g/2)t2+ v0sin&alpha; t

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

7. Oct 2, 2004

### 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

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

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

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

Last edited: Oct 2, 2004
8. Oct 2, 2004

### Tom McCurdy

Does this seem alright

9. Oct 2, 2004

### Pyrrhus

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

Last edited: Oct 3, 2004
10. Oct 4, 2004

### Tom McCurdy

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