Launch angle of projectile motion not in terms of time

Question

I have these two projectile-motion equations:

$$ vt\cos(\theta)=d $$

$$0=h+vt\sin(\theta)+\frac12at^2 $$

Is it possible to derive an expression for $\theta$ in terms of at most $d, v, h, a$ but not $t?$ The closest I've gotten is $$ t = \frac{d}{\cos(\theta)} $$ so $$ 0=h + \frac{dv\sin(\theta)}{v\cos(\theta)} + \frac12a\frac{d^2}{v^2 \cos^2(\theta)}\\ 0=h + d\tan(\theta) + \frac12a\frac{d^2}{v^2 \cos^2(\theta)}.$$

https://en.wikipedia.org/wiki/Projectile_motion#Angle_θ_required_to_hit_coordinate_(x,_y) contains a related result, but no derivation is given, and I am not sure how that result was attained.

Answer 1

Let $T=\tan\theta$, so your equation becomes $$0=h+dT+\frac{ad^2}{2v^2}(1+T^2)$$

You can solve this quadratic equation to get $\tan\theta$ in terms of $h,d,a$ and $v$

Answer 2

As you point out, for a projectile with an initial velocity $v_0$ and an angle of theta, the equations of motion give us $$y=-\frac{1}{2}gt^2+(v_0\sin{\theta})t$$ and $$x=(v_0\cos{\theta})t$$

As long as $\theta$ is less than $90^{\circ}$, $$\cos{\theta}\ne 0\implies t=\frac{x}{v_0\cos{\theta}}$$ and $$y=-\frac{1}{2}g\frac{x^2}{v_0^2\cos^2{\theta}}+x\tan{\theta}=-\frac{g}{2v_0^2\cos^2{\theta}}x^2 + x\tan{\theta}$$

Answer 3

You could try a substitution. For example, suppose $u = cos(\theta)$; then $\frac{sin(\theta)}{cos(\theta)} = \frac{\sqrt{1-u^2}}{u}$ and $\frac{1}{cos^2(\theta)} = \frac{1}{u^2}$. Your whole equation becomes the following:$$0=h+d\frac{\sqrt{1-u^2}}{u}+0.5a\frac{d^2}{u^2v^2}$$

Not what I'd call "pretty" by any stretch, but at least it's explicitly in only one unknown now. Clearing the denominators of $u$'s and solving for it yields:$$0=u^2h+du\sqrt{1-u^2}+\frac{0.5d^2a}{v^2}$$ $$-du\sqrt{1-u^2}=u^2h+\frac{0.5d^2a}{v^2}$$ $$u^2d^2(1-u^2)=(u^2h+\frac{0.5d^2a}{v^2})^2$$ $$-u^4d^2+u^2d^2=u^4h^2+\frac{u^2d^2ah}{v^2}+\frac{d^4a^2}{4v^4}$$

Let $z=u^2$. Rearranging to get everything on one side and making the substitution gets you the following:

$$(h^2+d^2)z^2+\frac{d^2(ah-v^2)}{v^2}z+\frac{d^4a^2}{4v^4}=0$$ $$z=\frac{\frac{d^2(v^2-ah)}{v^2}\pm\sqrt{\frac{d^4((ah-v^2)^2-a^2)}{v^4}}}{2(h^2+d^2)}$$ $$z=\frac{d^2((v^2-ah)\pm\sqrt{(ah-v^2)^2-a^2})}{2v^2(h^2+d^2)}$$

That's about as much as I have time for right now, I'm afraid, but you should get the gist by now. Presuming that the two values for $z$ are real, you'll be able to take the square root of the positive one to recover the values for $u$, and then take the arccos of those. Throw out whatever solutions don't correspond to the reality of the problem. As for finding $t$, all you need to do is compute $d/u$ for the values of $u$ you find; again, ignore any nonsensical results (like $t$ being negative for example).

Of course, all of this will only work if $(ah-v^2)^2-a^2 \ge 0$, which means that $(ah-v^2)^2 \ge a^2 \Rightarrow ah-v^2 \ge a \Rightarrow a(h-1)\ge v^2 \Rightarrow a\ge v^2/(h-1)$.

Addendum: taken to its ultimate conclusion, the values of $\theta$ can be computed with the given information using the following formula: $$\theta = \arccos(\pm\sqrt{\frac{d^2((v^2-ah)\pm\sqrt{(ah-v^2)^2-a^2})}{2v^2(h^2+d^2)}})$$

While the values of $t$ are given by:

$$t = \pm\frac{d}{\sqrt{\frac{d^2((v^2-ah)\pm\sqrt{(ah-v^2)^2-a^2})}{2v^2(h^2+d^2)}}}$$

I leave it to you to separate the valid cases from the invalid ones.