Does this example give a transcendental equation?

[kq6up]

I am teaching AP Physics, and wrote a problem on the board as an example of how to attack a problem. I new the answer would be nasty, but I didn't actually want them to solve it. I only wanted them to reduce the problem to one e.q. and one unknown. I was wondering if there is an analytic solution to the problem, or if it is transcendental.

The form it reduced to was 4=tan(th)*x-1/2*g*x^2*(sec(th))^2/vo^2

Here is the problem:

If a projectile is launched at 10m/s from the coordinates (0,0), what angle would it need to be launched at to pass through the points (4,4)?

Is it a transcendental equation, or is there a analytic solution?

Thanks,
Chris Maness

 

[CAF123]

Just to let you know I have done the algebra and arrived at an angle of ≈ 67^o.
This satisfies the original equation. I reduced it to:
$\frac{8g}{200} = 2\cosθ(\sinθ - \cosθ)$

 

[mfb]

There are two angles, close to each other, which satisfy the condition. If the point would be on the line given by 5-x^2/20 (using g=10), there would be one angle only. Above, it would be unreachable and below, there are two solutions.

Multiply the equation given in the first post by cos^2(θ)/4 and set x=4, g=10 and v=10 to get rid of constants and units:
$$\cos^2(\theta) = \sin(\theta)\cos(\theta) - \frac{1}{5}$$
This is similar to the equation in the second post.
As the angle is between 0 and 90°, sin(θ) is positive. Use $\sin(\theta)=\sqrt{1-\cos^2(\theta)}$, add 1/5 on both sides and multiply by 5:

$$5\cos^2(\theta) + 1 = 5\cos(\theta)\sqrt{1-\cos^2(\theta)}$$
Square both sides and substitute $\cos^2(\theta)=\frac{z}{5}$:

$$z^2 + 2z + 1 = z(5-z)$$
Simplify:
$$2z^2 - 3z + 1 = 0$$
Solve for z:
$$z=\frac{1}{4} ( 3 \pm 1)$$
This means
$$\cos^2(\theta_1)=\frac{1}{5}~~{\text{and}}~~\cos^2(\theta_2)=\frac{1}{10}$$
Which can be solved for the angles in an obvious way and gives about 63.4° and 71.6°.

 

[kq6up]

I am trying to solve it with Maxima, and it sure is giving me some strange results. The equation I posted above (I will give it a shot in LaTeX):

$\Delta y=\Delta x\cdot tan(\theta )-\frac { 1 }{ 2 } g{ \left( \frac { \Delta x }{ { v }_{ \circ }\cdot cos(\theta ) } \right) }^{ 2 }$

Does this look correct?

I tried plotting both sides to see where the intersect. It does make sense that there are two solutions. If the physics is good, I am going to try to post this equation in a Maxima forum to see why I can't get it to find roots for this. If I take my original equation and simplify it (as shown above by multiply through by cos^2(x)), I can plot both sides, works perfectly. I just can't get the original to work correctly.

Excuse me if I come off as a n00b. I haven't used my physics in a while (this is my first year teaching AP Physics).

Regards,
Chris Maness

 

[cabraham]

Is this the "field goal problem"? When launching a projectile from ground level, an angle of 45 degrees gives maximum distance if the final position is at ground level. If, however, we wish to get the maximum range at some altitude other than ground, I tried solving the problem in the 1970's & got a messy equation I didn't feel like tackling. It was only extra credit so I didn't do it. My physics prof never gave us the answer.

Claude

 

[nasu]

Is it a transcendental equation, or is there a analytic solution?

Yes, it is a transcendental equation. Even if you simplify it to "analytic" solutions like sin(x)=c, this will be just another transcendental equation, properly speaking.

transcendental equation - an equation that involves transcendental functions.
transcendentals functions: trigonometric functions, exponential, log, etc.

 

[Neandethal00]

Is this the "field goal problem"? When launching a projectile from ground level, an angle of 45 degrees gives maximum distance if the final position is at ground level. If, however, we wish to get the maximum range at some altitude other than ground, I tried solving the problem in the 1970's & got a messy equation I didn't feel like tackling. It was only extra credit so I didn't do it. My physics prof never gave us the answer.

Claude

When I used to solve this problem for Field Goal, I was also given the projectile Angle.
Then it is a piece of cake.

btw, Kg6Up's equation above is correct.
But I never tried a graphical solution. It would be interesting
to find one.
It seems simple, 2 equations with 2 unknowns.
But one equation becomes a quadratic equation of (time)²