# Projectile Motion - Elevated Target

1. Oct 8, 2005

### 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?

Matt

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2. Oct 8, 2005

### hotvette

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.

Last edited: Oct 8, 2005
3. Oct 8, 2005

### matrix0f8h

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.

4. Oct 8, 2005

### 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.

Last edited: Oct 8, 2005
5. Oct 8, 2005

### 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.

6. Oct 8, 2005

### 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.

Last edited: Oct 8, 2005
7. Oct 10, 2005

### 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.

Last edited: Oct 10, 2005
8. Oct 10, 2005

### matrix0f8h

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!

Last edited: Oct 10, 2005
9. Oct 10, 2005

### 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.

#### Attached Files:

• ###### projectile.gif
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Last edited: Oct 11, 2005
10. Oct 13, 2005

### 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.

11. Oct 13, 2005

### hotvette

It was fun. Thanks for the challenge!