Projectile Motion - Elevated Target

[matrix0f8h]

I was told by a professor that there was no analytic solution (it's a numerical methods course) when there is a elevated target. However, I can't find anything wrong with the analytic solution I came up with. Would anyone mind confirming that my formula (attached gif) for the launch angle when the initial velocity, range, target altitude are known is correct?


In addition, I understand that there are two angles for hitting the same target. One is the theta in the attached formula, the other is: PI/2 - THETA. I just don't know how to show this mathmatically. Any pointers? Is there something I am missing with the arcsin of a double angle?

Thanks in advance,
Matt

 

[hotvette]

In addition, I understand that there are two angles for hitting the same target. One is the theta in the attached formula, the other is: PI/2 - THETA. I just don't know how to show this mathmatically. Any pointers?

This is true only if x < x_{max}. Using two equations of motion for a projectile y(\theta,t) and x(\theta,t). You you can combine them to eliminate t, resulting in y(\theta,x). Solving this for \theta will result in 2 values.

Actually, I'd be willing to bet that if x = x_{max} the solution is two identical roots, \frac{\pi}{4}.

Re the first item, I didn't check your math, but there is no reason an analytical solution isn't possible for an elevated target. In fact, my Intro Dynamics book has a problem solution for kicking a football for a field goal, the goal post cross bar being an elevated target.

 

[matrix0f8h]

This is true only if x is less than the max x. Using two equations of motion for a projectile y(\theta,t) and x(\theta,t). You you can combine them to eliminate t, resulting in y(\theta,x). Solving this for \theta will result in 2 values.

I believe this is what I did in the attachment. Is it not? Yet, I only got one value... Not sure what I am missing.

Sorry it isn't in TEX. I did it in OpenOffice.org which is almost not completely TEX-like.

 

[Physics Monkey]

I have to agree with hotvette, it's certainly possible find an analytic solution. This is basic intro physics parabolic motion. Perhaps the professor was including air resistance? I don't know what he/she could have meant otherwise.

 

[hotvette]

Re two angles, I thought you were asking in general as opposed to an elevated target. The approach I mentioned works for a non-elevated target, but I'm not sure about elevated target. Hmmm, I need to think and reason a little.

 

[hotvette]

A couple of things seem strange in what you did. In (3), you eliminate \theta from the equation only to use it later in (4) to solve for \theta. It may be valid, I've just never seen a derivation like this before, especially having it in terms of v_{yf}. Secondly, in (3), you choose to use only the 2nd solution.

Think of this. Take (4) and solve for t. Substitute this result into (1). Thus, you'll end up with a quatratic equation in theta. If you know x & y, there should be two solutions for theta. The only catch is the x,y position of the target must lie inside the trajectory at \theta = \frac{\pi}{4}. Otherwise, there is only 1 solution or no solution.

I may see the numerical analysis part. If you do as I suggest, you end up with an equation with terms like (cos\theta)^2 and tan\theta. This may not be solveable analytically for aribtrary x & y.

Another thing I noticed. Your solution has a v_f term in it. In reality, v_f is a function of \theta. Think about it.

 

[hotvette]

I'm afraid some or most of what I said before is bunk. I've been playing with this for a while. Re 2 angles resulting in the same ending location (at least for non-elevated), the answer is quite simple, actually. Since we're dealing with periodic functions (sin, tan), multiple angles produce the same value of trig function. In the case of the simple trajectory, \theta_1 + \theta_2 = \frac{\pi}{2}. If \theta = \frac{\pi}{4}, then the 2 solutions are identical because \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.

If you do what I said before, that is, combine y(\theta,t) and x(\theta,t) into y(\theta,x), set y = 0 (completed trajectory for non elevated target), do some manipulation and using the trig identity sin(2\theta) = 2sin\theta cos\theta you get the followng:

sin(2\theta) = \frac{gx}{v_0^2}

As long as the right hand side is a valid value for sin(2\theta), there are 2 values of \theta that satisfy the equation. The same would be true for an elevated target, but the math is much more complex (I didn't go through it all), and I don't believe there is a closed form solution, thus the need for numerical analysis.

 

[matrix0f8h]

Another thing I noticed. Your solution has a v_f term in it. In reality, v_f is a function of \theta. Think about it.

Indeed it is. I probably should have not skipped so many steps in the orignal post... v_f = \sqrt{v_0^2 - 2gy_f}

Thanks for the replies. Still processing your later posts...

EDIT... What I said above didn't make much since... there was no theta... so you have something here... looking into it. Thanks!

 

[hotvette]

OK, finally got it. It was right in front of my eyes all along and I didn't see it. If you combine the equations for x(t) and y(t) into one equation to eliminate t, you get something that looks like:

y = \frac{a}{cos^2 \theta} \ + \ b \ tan \theta \ \Longleftrightarrow \ y = a \ sec^2 \theta \ + \ b \ tan \theta

Using the trig identity sec^2 = tan^2 + 1 you end up with an equation in tan \theta and tan^2 \theta, which is a quadratic equation in tan \theta that can be readily solved. Attached thumbnail illustrates the solution for specific values of x, y, v_0 that I chose.

The challenge question is to find the equation of the curve that bounds the valid values of x,y.

EDIT: I now follow what you did, and why you picked the second t. Everything you did was valid except getting from (4) to (5). First, there is a math mistake, plus you solve for \theta in terms of something that is a function of \theta. If you go from (3) to (4) w/o the last simplifying step in (3), you'll get an equation in sin \theta, cos \theta, and \sqrt{asin^2 \theta \ - \ b}. Solving this for \theta (looks complicated) should yield the same answer as the approach I took. All you did was solve for t first, then substitute, whereas I substituted to eliminate t, then solve for \theta. Both should yield the same result.

 

[matrix0f8h]

Thanks for all your help. Definately helped me sort through the solution.

I ended up implementing the numerical solution (just using the brute force method through all values of theta basically root finding). But it's nice to know that I wasn't completely off base with my assertion that there was an analytical solution.

 

[hotvette]

It was fun. Thanks for the challenge!