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I am trying to throw an object in my simulation with several criteria.

Object is thrown from [x0,y0]

object have to pass through point [x1,y1]

the top-most point is [m,n], where n - y1 = y1 - y0.

parabola image

So I have two points and top boundary of projectile motion.

I need to calculate, what linear velocity should I apply to my object in order to follow desired trajectory (no damping is taken into account).

I have tried to calculate it many times, using trajectory equations x=x0+v*t y=y0+v*t-0.5g*t*t and using parabola equation, but every time messed something up.

how can I calculate it?

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1 Answer 1

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To make things a bit simpler, I'll assume that $(x_0, y_0)=(0, 0)$. We can always translate things back once we're done. Also, to save writing subscripts I'll rename your $(x_1, y_1)$ to be $(a, b)$.

Describing the parabola.

The first thing we'll do is get the equation of the parabola in Cartesian coordinates. Once that's done we'll try to parametrize it as a trajectory, which of course is your ultimate goal. It's almost immediate that a downward-opening parabola through $(0,0)$ with apex coordinates $(m, n)$ is given by $$ y=n-(x-m)^2 $$ In this case, $n=2y_1$ (since $y_0=0$) which in our terms gives $n=2b$. Also, since the parabola contains the origin, we'll have $0=2b-(0-m)^2$, so $m=\sqrt{2b}$ so the parabola will be described by $$ y=2b-(x-\sqrt{2b})^2=2\sqrt{2b}\; x-x^2 $$ The apex will be at $(\sqrt{2b}, 2b)$. Note this, we'll use it shortly.

[Although we won't use it in what follows, we note that since the parabola contains $(a, b)$ it's not hard to show that $a=\sqrt{b}\;(\sqrt{2}-1)$.]

Parametrizing the parabola.

For initial velocity $V$ and elevation angle $\alpha$, the trajectory will be given by $$ \begin{align} x(t) &= V\cos(\alpha)\;t\\ y(t) &= V\sin(\alpha)\;t-\frac{1}{2}gt^2 \end{align} $$ For this trajectory, the apex will be when the derivative $y'(t_{apex})=0$. This gives us $$ t_{apex}=\frac{V}{g}\sin\alpha $$ which gives us $$ \begin{align} x(t_{apex})&=\frac{V^2}{g}\sin\alpha\cos\alpha\ \\ y(t_{apex})&=\frac{V^2}{g}\sin^2\alpha \end{align} $$ So the apex in these terms will be $((V^2/g)\sin\alpha\cos\alpha, (V^2/g)\sin^2\alpha)$. Now recall that the apex is $(\sqrt{2b}, 2b)$ so we'll have $$ \left(\frac{V^2}{g}\sin\alpha\cos\alpha\right)^2=\frac{V^2}{g}\sin^2\alpha $$ from which we find a relation between $V\text{ and }\alpha$: $$ V=\sqrt{\frac{g}{2}}\frac{1}{\cos\alpha} $$ Substituting this for $V$ in the original parametrized equations gives us $$ \begin{align} x(t) &= \sqrt\frac{g}{2}\;t\\ y(t) &= \sqrt\frac{g}{2}\;\tan(\alpha)\;t-\frac{g}{2}t^2 \end{align} $$ Now note that $x^2=(g/2)\;t^2$ and so we now have two equations for our curve, the parametrized version and the Cartesian version: $$ y=\tan(\alpha)\;x-x^2=2\sqrt{2b}\;x-x^2 $$ so we'll have $\tan\alpha = 2\sqrt{2b}$, so, finally, we have the elevation and the initial velocity given by $$ \alpha=\tan^{-1}(2\sqrt{2b})\qquad V=\sqrt{\frac{g}{2}(1+8b)} $$ (since, if $\alpha=\tan^{-1}(2\sqrt{2b})$, we'll have [Edit: $\cos\alpha=\underline{1/\sqrt{1+8b}}$]). These seem pretty reasonable, for example (in English units, rather than metric, since it makes the math a touch tidier),

  • If $b=1$ foot, then $V \approx 12$ ft/sec, max height = 2 feet, horizontal travel $\approx$ 2.8 feet
  • If $b=6$ feet, then $V \approx 28$ ft/sec, max height = 12 feet, horizontal travel $\approx$ 6.9 feet

That was fun. Thanks for posing the question.

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no no no, I thank you Rick! :) You had to put big effort in this! once again, thank you. I need to apply force by dimension, so I need to apply separately X and Y force. So when alpha is the angle I should apply the force, then cosα=1/(1+8b)=xForce right? And then from pythagoras theorem yForce = (g/2)*(1+8b)-xForce^2 rigt? –  relaxxx Oct 19 '12 at 20:04
TxForce is $V\cos\alpha$. We've established that $V=\sqrt{g(1+8b)/2}$ and that $\cos\alpha = 1/\sqrt{1+8b}$ so the xForce should be $\sqrt{g/2}$. yForce is $V\sin\alpha$ which turns out to be just $2\sqrt{gb}$, since $\sin\alpha=(2\sqrt{2b})/\sqrt{1+8b}$. –  Rick Decker Oct 20 '12 at 19:04

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