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Recently I have been working on the problem of calculating a ballistic trajectory. I landed on the following equation: $y = tan(θ)x - gt^2 $, where $θ$ is the launch angle, $x$ is the distance travelled horizontally, g is 9.81, t is time and y is the height relative to the starting position after having travelled x distance horizontally. However, the problem I am having is that as θ increases so does y after x distance travelled, which does not correctly model real-world behaviour. There should be a point at which launching the object at a higher angle decreases the height after x distance travelled. I have previously seen equations involving other trig functions when calculating a ballistic trajectory, but I wanted a good explanation as to why this is needed.

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    $\begingroup$ If you are given $x$ and wish to find the point $(x,y)$ on the trajectory, how do you get the value of $t$, which you also need in order to apply your formula? $\endgroup$
    – David K
    Aug 26, 2022 at 23:37
  • $\begingroup$ x/v where v is velocity $\endgroup$ Aug 26, 2022 at 23:38
  • $\begingroup$ Ok thanks I will put it up there $\endgroup$ Aug 26, 2022 at 23:39
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    $\begingroup$ Your error is that $t \neq x/v$ because the projectile is not just moving along the $x$ axis. $\endgroup$
    – David K
    Aug 26, 2022 at 23:40
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    $\begingroup$ I guess your equation came from $ y = vt \sin\theta - \frac 12 gt^2$, with $vt = x/\cos\theta$. If you actually substitute the remaining $t$ (in the $-\frac 12 gt^2$) too, you might get something that better link $x$ and $y$. $\endgroup$
    – peterwhy
    Aug 26, 2022 at 23:44

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With the given acceleration and initial speed and position: $$ \ddot r(t)=(0,-g)\\ \dot r_0=(v_0\cos(\theta),v_0\sin(\theta))\\ r_0=(0,0) $$ the integration gives: $$ \dot r(t)=(v_0\cos(\theta),v_0\sin(\theta)-gt) \\ r(t)=(v_0\cos(\theta)t,v_0\sin(\theta)t-\frac12 gt^2) $$ so the horizontal and vertical position are: $$ x(t)=v_0\cos(\theta)t\\ y(t)=v_0\sin(\theta)t-\frac12 gt^2 $$

And the last expression can be written as: $$ y(t)=\tan(\theta)x(t)-\frac12 gt^2 $$ or better, replacing the $t$: $$ y=\tan(\theta)x- \frac{g}{2v_0^2\cos(\theta)^2}x^2 $$

The last is a function for a parabola with negative convexity in the xy plane.

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