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Thread: Two answers to projectile problem?

  1. #1
    Junior Member B9766's Avatar
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    Two answers to projectile problem?

    During lecture, the professor provided an example of a projectile problem using trig.

    The problem is: A ball is hit at some angle $\theta$, and at it's initial speed upon being hit is 100ft/sec. It hits the ground 300 feet away. Given the formula:

    $\Large r\ =\ \frac{v_{0}^2}{32}\ sin(2\theta)\ \ where, r\ =\ 300\ and\ v_{0}\ =\ 100$

    What is the angle $\theta$?

    He goes on the solve the equation, ending up with:

    $\Large \theta\ =\ \frac{1}{2}sin^{-1}(\frac{24}{25})\ \approx\ 36.87^{\circ}$

    All good and fine. But then he says there are two answers:

    $\Large 36.87^{\circ}\ and\ \ 90^{\circ}-36.87^{\circ}\ =\ 53.13^{\circ} !!$

    He doesn't explain why he arrived at the second answer. The "How" is pretty obvious - he took the complement of the first answer. Can someone explain to me why the complement is also a valid answer?
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  2. #2
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    Re: Two answers to projectile problem?

    The equation

    $\displaystyle \sin (2t)=\frac{24}{25}$

    has infinitely many solutions

    If we know one solution, say $\displaystyle \alpha $,
    then all solutions are given by

    $\displaystyle 2t=\alpha + (360{}^{\circ})k$

    or

    $\displaystyle 2t=180{}^{\circ}-\alpha + (360{}^{\circ})k$

    If we take $\displaystyle k=0$ the first equation gives $\displaystyle t=\frac{\alpha }{2}$

    and the second one gives $\displaystyle t=90{}^{\circ}-\frac{\alpha }{2}$
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  3. #3
    Junior Member B9766's Avatar
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    Re: Two answers to projectile problem?

    Sorry Idea. I don't follow. It looks like you're substituting t for $\theta$ but I don't see how you got to $sin(2t)\ =\ \frac{24}{25}$ from $\theta\ =\ (\frac{sin^{-1}}{2})\frac{24}{25}$. Remember that I'm just learning the trig.
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: Two answers to projectile problem?

    Quote Originally Posted by B9766 View Post
    Sorry Idea. I don't follow. It looks like you're substituting t for $\theta$ but I don't see how you got to $sin(2t)\ =\ \frac{24}{25}$ from $\theta\ =\ (\frac{sin^{-1}}{2})\frac{24}{25}$. Remember that I'm just learning the trig.
    You are treating $\displaystyle sin^{-1}$ as if it is a variable. It is not. So it would be $\displaystyle \theta = \frac{1}{2} sin^{-1} \left ( \frac{24}{25} \right )$. Just like any other function $\displaystyle sin^{-1}(~)$ has to have something to operate on. In this case it's 24/25.

    So starting with
    $\displaystyle \theta = \frac{1}{2} sin^{-1} \left ( \frac{24}{25} \right )$

    $\displaystyle 2 \theta = sin^{-1} \left ( \frac{24}{25} \right )$

    $\displaystyle sin (2 \theta ) = sin \left [ sin^{-1} \left ( \frac{24}{25} \right ) \right ] = \frac{24}{25}$

    In this case we have to take a bit more care because of the domain of the sine and arcsine functions. $\displaystyle sin( \pi - \theta ) = sin(\pi)~cos(\theta) - sin(\theta) ~cos(\pi) = sin( \theta)$. This is the origin of the two solutions instead of just the one.

    -Dan
    Last edited by topsquark; Aug 6th 2018 at 02:59 PM.
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  5. #5
    Junior Member B9766's Avatar
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    Re: Two answers to projectile problem?

    Ah yes. That was a stupid shortcut I took putting $sin^{-1}$ above the 2 and not making $(\frac{24}{25})$ part of the arcsin function. Thanks for the help with the sin of the arcsin. I know we studied that but it certainly didn't stick. I do understand your answer.

    And I understand Idea's statement that the answer occurs every $2\pi$ cycle such that $2\theta\ =\ \alpha\ +\ 2\pi n$. But I don't understand the second statement that $2\theta\ =\ \pi\ -\ \alpha\ +\ 2\pi n$. I think that was really my original question. Where does the extra $180^{\circ}$ come from?
    Last edited by B9766; Aug 6th 2018 at 05:36 PM.
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Two answers to projectile problem?

    It can be shown that the trajectory of a projectile where gravity is assume constant, and drag ignored, will take the form:

    $\displaystyle y(x)=\tan(\theta)x-\frac{g}{2v_0^2\cos^2(\theta)}x^2$

    The coordinate axes have been oriented such that the origin coincides with the projectile's initial position.

    Equating $\displaystyle y$ to zero and taking the non-zero root, we get the range $\displaystyle r$ of the projectile as:

    $\displaystyle r=\frac{v_0^2\sin(2\theta)}{g}$

    We see now that the range is a maximum for $\displaystyle \theta=\frac{\pi}{4}$ and is given by:

    $\displaystyle r_{\max}=\frac{v_0^2}{g}$

    But, for any range $\displaystyle 0<r<r_{\max}$ (with everything else being the same), there are two launch angles that will give us the same range $\displaystyle r$, although the time spent to get there will be different for the two resulting trajectories.

    See the attached diagram below, which for some reason I was unable to attach inline.
    Attached Thumbnails Attached Thumbnails Two answers to projectile problem?-mhf_0005.png  
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    Re: Two answers to projectile problem?

    Quote Originally Posted by B9766 View Post
    Ah yes. That was a stupid shortcut I took putting $sin^{-1}$ above the 2 and not making $(\frac{24}{25})$ part of the arcsin function. Thanks for the help with the sin of the arcsin. I know we studied that but it certainly didn't stick. I do understand your answer.

    And I understand Idea's statement that the answer occurs every $2\pi$ cycle such that $2\theta\ =\ \alpha\ +\ 2\pi n$. But I don't understand the second statement that $2\theta\ =\ \pi\ -\ \alpha\ +\ 2\pi n$. I think that was really my original question. Where does the extra $180^{\circ}$ come from?
    Let's consider the sine function from $-\pi$ to $\pi$. It turns out that sine is an odd function (not odd as in peculiar, but odd as in odd vs even). This means that $\sin(-x) = -\sin x$. If you look at the graph of the function, it looks like a reflection across both the x- and y- axes at the same time. So, if we know what is happening on $[0,\pi]$, we can figure out what is happening on $[-\pi,0]$ by simply taking the negative. And because the function is periodic, if we understand how it operates over a single period, we know how it works everywhere. (The graph is going to be a translated copy of what we already know for $(2n-1)\pi \le x \le (2n+1)\pi$).

    We know that $\sin \pi = \sin 0 = 0$. We also know that $\sin \dfrac{\pi}{2} = 1$. And if we look at the sine function where $x$ goes from $0$ to $\dfrac{\pi}{2}$, it is a mirror image of the sine function where $x$ goes from $\dfrac{\pi}{2}$ to $\pi$. In other words, $\sin\left(\dfrac{\pi}{2}-x\right) = \sin\left(\dfrac{\pi}{2}+x\right)$. Similarly, we can start from $0$ and go to the right is the same as starting from $\pi$ and moving to the left. So, we have:

    $$\sin (0+x) = \sin(\pi-x)$$

    This is the formula that Idea used. The LHS simplifies to $\sin x$.
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  8. #8
    Junior Member B9766's Avatar
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    Re: Two answers to projectile problem?

    Thank you SlipEternal. Before I read your answer I graphed $y=sin(2\theta)$ and $y=0.5$ and could see the two values of y that satisfied the equation. So, intuitively I can see that there were two answers.

    Two answers to projectile problem?-5b6a8af3.jpg

    And I follow the two equations: $sin(\frac{\pi}{2}−x)=sin(\frac{\pi}{2}+x)$ and $sin(0+x)=sin(\pi−x)$ But somewhere I'm missing the connection to why the second point, $\theta_2 = \large \frac{\pi}{2}-\frac{arcsin(\frac{24}{25})}{2}$ uses $\frac{\pi}{2}$ rather than $\pi$..

    Just to make sure I didn't make another dumb mistake with my LaTex commands, I solved the original equation as follows:

    $300ft=\large \frac{100^2 sin(2\theta_1)}{32}$

    $sin(2\theta_1)=\large \frac{32*300}{100^2}$

    $arcsin(sin(2\theta_1))=arcsin(\frac{24}{25})$

    $2\theta_1=arcsin(\frac{24}{25})$

    $\theta_1= \large \frac {arcsin(\frac{24}{25})}{2}\normalsize \approx 36.87^\circ$

    Then, $\theta_2=\frac{\pi}{2}-36.87=53.13^\circ$


    Why was $\large \frac{\pi}{2}$ used instead of just $ \pi$ in the last step? Is it that the original equation specified $sin(2\theta)$ and that means the period is $\pi$ rather than $2\pi$, therefore the only other positive answer has to be $\theta_2 = \large \frac{\pi}{2} \normalsize - \theta_1$?
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  9. #9
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    Re: Two answers to projectile problem?

    When you divide $2\theta$ by 2, you divide everything on the RHS by 2, as well. Including the $\pi$.

    $$\pi-2\theta=\arcsin\left( \dfrac{24}{25}\right)$$

    $$2\theta = \pi - \arcsin\left( \dfrac{24}{25}\right)$$

    $$\theta = \dfrac{\pi}{2} - \dfrac{\arcsin\left( \dfrac{24}{25}\right)}{2}$$
    Last edited by SlipEternal; Aug 8th 2018 at 12:59 PM.
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  10. #10
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    Re: Two answers to projectile problem?

    Quote Originally Posted by B9766 View Post
    $\theta_1= \large \frac {arcsin(\frac{24}{25})}{2}\normalsize \approx 36.87^\circ$

    Then, $\theta_2=\frac{\pi}{2}-36.87=53.13^\circ$
    Ouch! Pick either radians or degrees and stick with it. $\displaystyle \frac{\pi}{2} - 36.87^{o}$ makes no sense.

    -Dan
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  11. #11
    Junior Member B9766's Avatar
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    Re: Two answers to projectile problem?

    Thanks Dan. There's a lot I'm trying to make sense of. All advice helps.
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  12. #12
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    Re: Two answers to projectile problem?

    SlipEternal and TopsQuark:

    First, thanks again for sticking with me on this. I think I finally get it. A big part of my initial problem was the inability to identify exactly where the disconnects existed in my mind. I'm only part way through this pre-calc class and don't have the benefit of learning all the trig covered in it yet. For example, the next chapter covers sum and difference formulas - so I'm not there yet.

    I was going to go into detail about what finally made it all gel for me but it came down to graphing the curves, sine values and actual angles in their quadrants, plus your equations and prompting. Once I was able to visualize the relationships of the angles to the sine function, I could see the $\displaystyle sin(\theta) = sin(\pi/2-\theta)$ equivalence.

    Thank you.
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    Re: Two answers to projectile problem?

    Quote Originally Posted by B9766 View Post
    SlipEternal and TopsQuark:

    First, thanks again for sticking with me on this. I think I finally get it. A big part of my initial problem was the inability to identify exactly where the disconnects existed in my mind. I'm only part way through this pre-calc class and don't have the benefit of learning all the trig covered in it yet. For example, the next chapter covers sum and difference formulas - so I'm not there yet.

    I was going to go into detail about what finally made it all gel for me but it came down to graphing the curves, sine values and actual angles in their quadrants, plus your equations and prompting. Once I was able to visualize the relationships of the angles to the sine function, I could see the $\displaystyle sin(\theta) = sin(\pi/2-\theta)$ equivalence.

    Thank you.
    No problem. To clarify, it is $\sin \theta = \sin (\pi-\theta)$. Or $\sin \theta = \cos \left(\dfrac{\pi}{2}-\theta \right)$.
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  14. #14
    Junior Member B9766's Avatar
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    Re: Two answers to projectile problem?

    SlipEternal:

    Kind of a stupid comment I made. Wasn't it?

    What I meant to say was that I understand why, in the equation that contains $sin(2\theta)$, the second angle $\theta_2$ can be computed from $90^\circ-\theta_1$ and why $90^\circ$ is used rather than $180^\circ$.

    Knowing $sin(\theta)=sin(\pi-\theta)$ is what threw me in the first place. I couldn't reconcile the use of $90^\circ$ in the former with $\pi$ in the latter.

    It made a lot more sense to me when I graphed $sin(2\theta)$ with intersecting lines at $\theta_1=36.87^\circ$ and $\theta_2=53.13^\circ$, and understanding your explanation of $sin \left(\dfrac{\pi}{2}-x\right)=sin\left(\dfrac{\pi}{2}+x\right)$.

    I hope that made a lot more sense.
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