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Using this equation I can get the desired launch angle of a projectile, where $\theta$ = the angle, $g$ = gravity, $x = x$ coordinate of impact point, and $y = y$ coordinate of impact point.

$\theta =\arctan {\left({\frac {v^{2}\pm {\sqrt {v^{4}-g(gx^{2}+2yv^{2})}}}{gx}}\right)}$

Assuming a time scale(t) of $0.0 - 1.0, 0.0$ being the launch of the projectile and $1.0$ being the impact, how could I get the coordinate pair in 2D space for the position of the projectile at any given t between $0.0$ and $1.0$? Not a mathematician or a physicist, but do have a basic understanding of algebra and trig.

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  • $\begingroup$ Where did you get your equation from? $\endgroup$ – sammy gerbil Mar 29 '17 at 3:13
  • $\begingroup$ en.wikipedia.org/wiki/Trajectory_of_a_projectile $\endgroup$ – GeoJohn Mar 29 '17 at 3:59
  • $\begingroup$ The equations quoted by Feynman immediately follow the equation you have quoted in the wiki article. ... In the title you ask for position as a function of time but in the text you ask for (x,y) pairs. That is given by the 3rd eqn after yours. $\endgroup$ – sammy gerbil Mar 29 '17 at 4:03
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Firstly, construct a reference frame and $y$ is upward. Secondly, list motion equation. $$m\ddot{x}=0$$ $$m\ddot{y}=-mg$$ Thirdly, solve motion equation. $$x(t)=v_0^xt+x_0$$ $$y(t)=-\frac{1}{2}gt^2+v_0^yt+y_0$$ in which $v_0^x,x_0,v_0^y,y_0$ are initial conditions at time $t=0$.

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    $\begingroup$ For completeness, the angle $\theta$ of the OP enters as $v^y_0/v^x_0=\tan(\theta))$.. With the time-of-flight $=1$ and the horizontal distance known, one gets $v_0^x$ and from there the $y$-component. $\endgroup$ – ZeroTheHero Mar 29 '17 at 2:57
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In general, to solve for position at some time, you could use this process:

Split everything into two components: horizontal and vertical. First, find the initial velocity components. Then calculate position for each component.

Horizontal is easy, assuming no forces act on the projectile such as wind resistance. The velocity remains constant. So just calculate the initial horizontal velocity. Given the initial horizontal position, the horizontal position at t is velocity times t plus original position.

Vertical is not much harder: velocity changes at a constant rate of acceleration, g. Use SUVAT equations like above, given your initial vertical velocity component, u.

You need one extra step: find time of impact, and "normalise" above equations so start time is 0 and end is 1. When the projectile hits the ground, vertical displacement is zero; so using your SUVAT equation, solve to find the two values of $t $ that satisfy it. Obviously you want the greater value, since the first will be the launch.

Initial velocity components

Given angle theta, calculate horizontal and vertical velocity components:

$$ v_x = v \sin \left(\theta\right) $$

$$ v_y = v \cos \left(\theta\right) $$

A right-angled triangle could be drawn to depict the relationship. $v $, $v_x $ and $v_y $ form the sides, with $\theta $ being the angle between $v $ and $v_x $.

From there

Feynman beat me to it with his very succinct answer; perhaps mine gives a helpful introduction to how his formulae could be obtained.

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  • $\begingroup$ Indeed. The initial velocity components and explanation there of, help much in relating back to Feynman's answer. +1 for this, I can't except both, but would if I could. $\endgroup$ – GeoJohn Mar 29 '17 at 4:21

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