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High school students are learning about the basics of solving quadratics and trigonometric ratios, including trigonometric inverses. The eventual goal of their project is to be able to show a reasonable firing solution, given in initial angle $\theta$ and initial velocity $v_0$.

Projectile motion is given by

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

where $x$ and $y$ are horizontal and vertical displacement. In this equation, one might change $\theta$ or $v_0$ and the equation behaves as expected - allowing students to directly give a reasonable solution.

My goal is to find a way to help my students build up understanding from basic quadratics, which appear like a trajectory in shape, to a more accurate representation of trajectory given by the equation above. Since the trajectory equation is quadratic in nature and can be reduced to $ax+bx^2$, and because students will also learn trigonometric ratios, this is reasonable. Note that I am giving the context because the answer needs to be at a relatively low level, and I also hope that someone might have a contribution for the approach.

Students initially create quadratics that look like projectile motion using quadratics of the form $y=-x(x-a)$. This gives a nice way to discuss solutions (the impact point) using the zero product property. We build up to $ax^2 +bx+c$ form, now discussing other ways to find solutions. Then I ask students to convert this equation to an initial angle and velocity (after discussion of what a firing solution might constitute). They realize quickly that such an equation doesn't make a lot of necessary information very obvious - eg, gravity, angle, velocity, etc.

While we eventually discuss the trajectory motion equation (sadly, I haven't discovered how to build this in a way high school geometry students understand), we initially account for gravity using the equation

$$h(t)=\frac{1}{2}gt^2+v_{0,y} t+h_0$$

where $t$ is time in seconds, $g$ is acceleration due to gravity, and $h(t)$ is height in meters, and $v_{0,y}$ is initial $y$ velocity. My question is: can $h(t)$ be modified with a particular initial angle $\theta$ and initial velocity $v_0$? My suspicion is that the answer to this question will look like the parameterized version of the trajectory equation.

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  • $\begingroup$ Which equation do you mean by "this equation"? $\endgroup$ – edm Jun 13 '17 at 14:13
  • $\begingroup$ Edited for clarity - I meant the equation most proximal - $h(t)$ $\endgroup$ – Zediiiii Jun 13 '17 at 14:50
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Here's how I understood this before I studied calculus:

Suppose the projectile is to be launched from the point $(0,0)$ at initial speed $v_0$ in a direction making an angle $\theta$ above the positive $x$ axis, and suppose (for now) that there is no gravity here. Then the projectile travels a distance $v_0$ in that direction in the first unit of time, and a distance $v_0 t$ in that direction in the first $t$ units of time.

Construct a right triangle with legs parallel to the $x$ and $y$ axes and hypotenuse $v_0 t.$ The horizontal leg is then $v_0 t\cos\theta$ and the vertical leg is $v_0 t\sin\theta.$ If we construct this triangle on the coordinate plane with the hypotenuse in the initial direction of the projectile with one end of the hypotenuse at $(0,0),$ we find that the $(x,y)$ coordinates at time $t$ are given by \begin{align} x &= v_0 t\cos\theta,\\ y &= v_0 t\sin\theta & \text{(in zero gravity).} \end{align}

Now we consider the effects of gravity. Assuming you have already derived the fact that an object starting at $(0,0)$ with no initial velocity will fall under the influence of gravity so that it reaches the vertical coordinate $y = -\frac12 g t^2$ at time $t,$ we combine the motion due to the initial velocity of the projectile with the motion due to gravity. The result is that the projectile ends up at a position displaced vertically by $-\frac12 g t^2$ from where it would have been if there were no gravity, and not displaced horizontally (because things don't fall sideways!), so it ends up at \begin{align} x &= v_0 t\cos\theta, \tag1 \\ y &= v_0 t\sin\theta - \tfrac12 g t^2 & \text{(where gravity is $g$).}\tag2 \end{align}

To really understand this, I think you need some notion of Galilean relativity: the launch speed and angle establish an inertial frame of reference traveling at velocity $v_x=v_0\cos\theta,$ $v_y=v_0\sin\theta$ relative to the ground, and the projectile falls straight down within that frame of reference. One way to explain it might be, you fire two cannonballs from the same place in the same direction at the same speed; one is a magic cannonball that isn't affected by gravity, but the other is an ordinary cannonball. A hypothetical observer riding on the magic cannonball will initially see the ground receding down and behind while the ordinary cannonball is stationary, but after $t=0$ the ordinary cannonball will appear to fall straight down (as seen from the vantage point on the magic cannonball). So while the magic cannonball follows a trajectory with $y=v_0t\sin\theta,$ the ordinary cannonball follows Equation $(2).$

If you launch the projectile from a point $(x_0,y_0)$ instead of $(0,0),$ everything is displaced by $x_0$ horizontally and $y_0$ vertically, and you get the more general equations \begin{align} x &= v_0 t\cos\theta + x_0,\\ y &= v_0 t\sin\theta - \tfrac12 g t^2 + y_0. \end{align}

But as long as you launch from $(0,0),$ then $x_0 = y_0 = 0$ and the equations end up acting like $(1)$ and $(2).$

Now observe that if $x = v_0 t\cos\theta$ (from Equation $(1)$), then $$t = \frac{x}{v_0\cos\theta}.$$ Make this substitution for $t$ in Equation $(2)$ and you can work out the equation for $y$ as a function of $x.$

I don't agree with the way the question implies that parabolas of the form $y = ax + bx^2$ merely appear to be like trajectories in shape, nor the implication that the formula with $v_0$ and $\theta$ is more accurate than $y = ax + bx^2.$ On the contrary, I would say that $y = ax + bx^2$ is just as accurate an equation of a projectile trajectory (provided that $b$ is negative), even though it does not explicitly tell you what the initial velocity and angle are. If you want the initial velocity and angle of the projectile that follows the trajectory $y = ax + bx^2,$ you have to solve for $v_0$ and $\theta$ in terms of $a,$ $b,$ and $g.$


Update: For reference, it may be helpful to show the derivation of $v_0$ and $\theta$ from $a$ and $b.$ We suppose that we are given the path $$ y = ax + bx^2 $$ for some given (known) values of $a$ and $b.$ This can only be a projectile trajectory if the same path is also given by an equation of the form $$ y = (\tan\theta)x - \left(\frac{g}{2v_0^2\cos^2\theta}\right)x^2$$ for some values of $v_0,$ $\theta,$ and $g.$

In order for both equations to describe the same path, the right hand side must be the same polynomial in each equation. This means the coefficient of $x$ must be the same each time, and the coefficient of $x^2$ must be the same. So we have a projectile trajectory only if we can simultaneously solve these two equations: \begin{align} a &= \tan\theta, \\ b &= -\frac{g}{2v_0^2\cos^2\theta} = -\frac{g}{2v_0^2}\sec^2\theta. \end{align}

From the first of these equations we have $\theta = \tan^{-1}(a)$ (for a projectile that travels over the part of the parabola for which $x \geq 0$) or $\theta = \tan^{-1}(a) + \pi$ (for the other part of the parabola). Using the identity $\sec^2\theta = 1 + \tan^2\theta,$ we have $$ b = -\frac{g}{2v_0^2}\sec^2\theta = -\frac{g}{2v_0^2}(1 + \tan^2\theta), $$ and using the fact that $a = \tan\theta,$ $$ b = -\frac{g}{2v_0^2}(1 +a^2). $$

If $g$ is known, then there is only one unknown in this last equation, $v_0.$ To find $v_0$ we first isolate $v_0^2$ on one side of the equation: $$ v_0^2 = -\frac{g(1 + a^2)}{2b}. $$

The left side must be positive; making the usual assumption that $g > 0,$ and and noticing that $1 + a^2$ is necessarily positive, we conclude that we can only solve this equation if $b < 0.$ Provided that $b < 0,$ we find that $$ v_0 = \sqrt{-\frac{g(1 + a^2)}{2b}}, $$ so we have derived both the angle and initial velocity from the coefficients $a$ and $b.$ Moreover, this derivation always works as long as $b < 0$: any downward-opening parabola is the path of a projectile trajectory.

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  • $\begingroup$ Good point about trajectory - what I really meant was that if a student uses $ax+bx^2$ where $a$ and $b$ are arbitrary numbers, then their understanding isn't as accurate as if they deliberately created those two constants using understanding of velocity and angle. Further, students will be unable to easily derive $v_0$ and $\theta$ if $a$ and $b$ are arbitrary. I'd fix the question, but I'll leave it for others to learn from my misconceptions. =) $\endgroup$ – Zediiiii Jun 13 '17 at 18:39
  • $\begingroup$ That said, gravity needs to be accounted for in a specific way. While a student might be able to provide an acceptable looking trajectory with an arbitrary $a$ and negative $b$, won't that equation most likely fail to account correctly for gravity? $\endgroup$ – Zediiiii Jun 13 '17 at 18:44
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    $\begingroup$ For a given value of $g,$ there are only very specific sets of equations that correctly relate all three of the variables $x,$ $y,$ and $t.$ If you're only looking at the path in the $x,y$ plane ($y$ as a function of $x$), however, you just need to make sure the curvature of the parabola is consistent with the direction of gravity; that is, you need $b<0.$ As long as $b<0,$ you can always find an angle and initial speed that will follow the parabola $y = ax + bx^2.$ I've added the explicit derivation to the answer. $\endgroup$ – David K Jun 13 '17 at 20:26
  • $\begingroup$ For the record, this explanation went over extremely well. I added the trig connection that the coefficient of the $x$ term, $\tan \theta$, is the slope of the initial angle, so naturally, the inverse tangent of that coefficient provides the initial angle. Finding $v_0$ in terms of $a$ and $b$ is a great idea (why didn't I think of that?). My geometry kiddos have to take my word for it on that equation though. =) Excellent explanation. $\endgroup$ – Zediiiii Apr 4 '18 at 3:29
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You want $\boxed{h(t)=gt^2+v_0\sin(\theta)t+h_0}$.

This is because the vertical component of the $v_0$ arrow pointed in the $\theta$ direction has height $v_0\sin(\theta)$.

By the way, in this problem $g=-4.9$, because the acceleration due to gravity is $-9.81$ m/$s^2$.

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