Roots of transcendental equations in physics problem

Question

I have a physics problem where I have to find for which angle and time is the target at distance $r$ and height $0$ hit if initial velocity is $v_0$. I am aware that this problem is very basic and I also understand the physics behind it and how you solve it, but I am not sure about the mathematical side. So I have a set of equations:$$v_0\cdot \cos(x)\cdot t=r$$ $$v_0 \cdot \sin(x)\cdot t -\frac{gt^2}{2}=0$$ $$0<x<\pi/2$$ I know intuitively that I should have two angles and two times that solve the set of equations, but I am not sure how can one tell, mathematically speaking, that the given system has two solutions. I would normally solve this by factoring out $t$, but than I only get one valid solution. Only after solving $\cos(x)^2+\sin(x)^2=1$, I get both solutions.

Answer 1

The second equation $v_0 \cdot sin(x)\cdot t -\dfrac{gt^2}{2}=0$ says that $$(v_0 \cdot sin(x) - \frac{gt}{2})\cdot t = 0 .$$ Thus either $t = 0$ or $t = \dfrac{2v_0\sin(x)}{g}$. The solution $t = 0$ disqualifies because the first equation would produce $0 = r$. Therefore we must have $$t = \frac{2v_0\sin(x)}{g} .$$ Inserting in the first equation produces $$\sin(2x) = 2\cos(x)\sin(x) = \frac{gr}{v_0^2} . $$ Thus a solution $x$ exists if and only if $\dfrac{gr}{v_0^2} \le 1$, i.e $v_0 \ge \sqrt{gr}$. The third inequality says that $0 < 2x < \pi$, thus you get two solutions for $x$ if $v_0 > \sqrt{gr}$ and one solution $x = \pi/4$ if $v_0 = \sqrt{gr}$. Thus the final result is this:

Your system of three equations has a solution if and only if $$v_0 \ge \sqrt{gr} .$$ If $v_0 > \sqrt{gr}$ you get two angles $x \in (0,\pi/2)$ determined by $$\sin(2x) = \frac{gr}{v_0^2} .$$ The smaller angle $x_s$ has the property $0 < x_s < \pi/4$, the bigger angle $x_b$ has the property $\pi/4 < x_b < \pi/2$ and is related to $x_s$ by $x_b = \pi/2 - x_s$.

If $v_0 = \sqrt{gr}$ you get one angle $x = \pi/4$.

For each possible angle $x$ you get the time $$t = \frac{2v_0\sin(x)}{g} .$$

Edited:

The physical problem is to study a projectile shot at an angle $x \in (0,\pi/2)$ with the initial velocity of $v_0$, with gravity $g$ acting on it, air resistance is neglected. The shooting takes place in a plane.

Its trajectory lies an a parabola and has parameterization $$\phi(t) = (r(t),h(t)) = (v_0 \cdot \cos(x)\cdot t, v_0 \cdot sin(x)\cdot t -\dfrac{gt^2}{2}) \in \mathbb R^2 .$$ Therefore, given $x$ and $v_0$, the projectile hits the ground at the distance $r(t^*)$ where $h(t^*) = 0$ with $t^* > 0$. We get $t^* = \dfrac{2v_0\sin(x)}{g}$ and $$r(t^*) = \dfrac{v_0^2 \sin(2x)}{g} .$$ Clearly the maximal firing range occurs at an angle of $x = \pi/4$ where $\sin(2x) = 1$. Moreover, given $x$ and $r = r(t^*)$, we see that we must have $$v_0 = v_0(r) = \sqrt{\frac{gr}{\sin(2x)}} \ge \sqrt{gr} .$$