Optimizing Projectile Distance from a Plateau: A Challenging Kinematics Problem

[Machinus]

challenging kinematics problem

Here is a problem I have been trying in Maple, but to no avail.

I am trying to calculate the theta that will result in maximum distance for a projectile fired from off the edge of plateau onto another, lower surface. The projectile can be considered to be fired from the exact corner of the ledge so there is no minumim distance; a velocity of zero will result in the projectile moving no distance but falling onto the lower surface.

...v[0]
______/
...|
...|______


(ignore the periods. the slash is the vector)

The method I have tried is to try to solve for t using the y-coordinates of motion, and then plug this back into the x-equation, which is linear. However, in trying to solve for t with the y-equations, I get two very nasty solutions (due to the quadratic solution) that I really am not going to do by hand, and Maple has decided not to cooperate. I am using Maple 9.5 and I know my way around, and I don't think there is any more I can do this way. If anyone knows of a better way to solve this, please let me know. Thanks.

 

[chroot]

The angle for best range is always 45 degrees above the horizontal.

See this page: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra12

The range of a projectile is

[tex]R = \frac{v_0^2 \sin{2 \theta}}{g}[/tex]

Assuming [itex]v_0[/itex] is a constant, you're just solving for the maximum of [itex]\sin{2 \theta}[/itex]. The maximum value of the sine function is one, when its argument is [itex]\pi / 2[/itex], or when [itex]\theta = \pi / 4 = 45^o[/itex].

- Warren

 

[chroot]

Ah, actually, it serves me right for not reading carefully enough. You're trying to solve for the case with a launch platform higher than the landing site. (cyrus, you're right, it does indeed affect the best angle). You're trying to sovle the general equation shown here:

http://hyperphysics.phy-astr.gsu.edu/hbase/imgmec/tra9.gif

Sorry about that.

You should be able to just take the derivative of the expression with respect to theta and solve for its zeroes:

[tex]R = v\cos(\theta) \left[ \frac{2 v \sin(\theta) y}{g} \pm \sqrt{ \frac{(v \sin(\theta))^2}{g^2} - \frac{2 y}{g}} \right][/tex]

- Warren

 

[Machinus]

I was really distraught when you posted that! Thanks for the correction.

I did take the derivative and solve for zero. That proved to be quite unwieldy and difficult to simplify. Thank you for the confirmation, however. Is there any other way to solve this?

 

[chroot]

Machinus,

I'll put it through Mathematica when I get home. I agree in that I'm not about to try to solve that one by hand.

- Warren

 

[pervect]

OK, it takes some fooling around, but my ancient version of maple gets it. First we solve for the time as a function of the height h, assumed here to be positve when the projectile drops

[tex]
{\frac {{\it vy}+\sqrt {{{\it vy}}^{2}+2\,gh}}{g}}
[/tex]

then xmax is given by

[tex]
{\frac {v\cos \left( \theta \right) \left( v\sin \left( \theta
\right) +\sqrt {{v}^{2} \left( \sin \left( \theta \right) \right) ^{
2}+2\,gh} \right) }{g}}[/tex]

then we make the substitutions

sin(theta) = tan/sqrt(1+tan^2), cos(theta)=1/sqrt(1+tan^2)

and tell maple to simplify, we get

[tex]
{\frac {v \left( v \left( \tan \right) +\sqrt {{v}^{2}{\tan}^{2}+2\,gh
+2\,gh{\tan}^{2}} \right) }{ \left( 1+{\tan}^{2} \right) g}}
[/tex]

solving for the derivative of the above to be zero wrt tan, we get

tan = v/sqrt(v^2+2gh)

which is the value of the tangent of the angle theta.

 

[Machinus]

How did you get such a simple t? Also, aren't there two solutions?

 

[Cyrus]

hey warren, that website is incorrect. It should be Vy/g, not 2Vy/g I think they made a mathematical error. I got the same equation as pervect. Yes, two solutions, but you want the one that is positive and largest. This will be the max time. so the distance x=vx *t will have the largest value.

 

[pervect]

How did you get such a simple t? Also, aren't there two solutions?

I put

[edit] minus ! [edit]

y := vy*t - .5*g*t^2;
solve(y=-h,t);

into Maple. I also did a foo:=sub(t=<expr>,y);simplify(foo);

But I''d encourage you to double check, just on general principles.

 

[chroot]

pervect:

Something about your tans without arguments is deeply unsettling to me.

- Warren

 

[Cyrus]

hey warren, did u see the error?

 

[pervect]

pervect:

Something about your tans without arguments is deeply unsettling to me.

- Warren

OK, use 'u' or something that's more innocuous if 'tan' without arguments is too unsettling...

Also, aren't there two solutions?

I forgot to mention - choose the time solution that's positive.

 

[Machinus]

Well, thanks to all for the help, but my professor from Fermi has solved it for me relatively easily. I haven't seen his work yet but he presented the solution to another student who knew that I was working on the problem. Without further ado:

[tex]\theta_{max}=\arctan(\frac{v_{0}}{\sqrt{v_{0}^2+2gh}})[/tex]

If you will note, an encouraging check to this problem is to substitute zero for the height, and the formula reduces to

[tex]\theta_{max}=\arctan(\frac{v_{0}}{v_{0}})[/tex]

which is of course 45 degrees, the correct value from a projectile on a flat surface.

Thank you professor Davenport!

 

[Cyrus]

We also got that anwser machinus, did you read what pervect posted, he solved it with maple

solving for the derivative of the above to be zero wrt tan, we get

tan = v/sqrt(v^2+2gh)

 

[Machinus]

Ah, yes, now I see that they are the same answer. I thought it was kind of unclear, but you are right. Thank you!

 

[Cyrus]

Dont thank me thank pervect, and I will thank him too, thanks. I thought that was an interesting problem. What did the angle you were looking for turn out to be?

 

[Machinus]

The angle is dependent on the height. Right now I am working on making a nice-looking graph to display the results.