Inequality from a projectile motion problem

Question

Imagine this scenario: Two objects are thrown simultaneously, from $y=0$, with different speeds $u$ and $v$, and at angles $\varphi$ and $\varphi+\theta$. We want to find out which projectile returns to its original position first (i.e $y=0$). When they are both back at it, the distance between them is $\Delta x$, which we assume is positive.

We assume that $\varphi+\theta\leq\pi/2$, and that the angle $\varphi$ is associated to the projectile thrown with speed $u$.

Given: $\varphi+\theta\leq\frac{\pi}{2}$, $\Delta x\geq0$.

For both projectiles, we can express their motion vector like this: $$\vec r(t)=(v_xt)\hat\imath+(v_yt-\frac{1}{2}gt^2)\hat\jmath$$ If I suppose that $A$'s final position is $x$, then I can find the time like this: $$t_A=\frac{x}{v\cos(\varphi+\theta)}=\frac{2v}{g}\sin(\varphi+\theta)$$ Doing the same thing for $B$, I get: $$t_B=\frac{x+\Delta x}{u\cos(\varphi)}=\frac{2u}{g}\sin(\varphi)$$ Taking the difference using the solution from the $y$ coordinates: $$t_A-t_B=\frac{2}{g}(v\sin(\varphi+\theta)-u\sin(\varphi))$$ I know that $\sin(\varphi+\theta)\geq\sin(\varphi)$, but I have no idea how the velocities are. The quotient of the times can help me, though: $$\frac{t_A}{t_B}=\frac{v}{u}\frac{\sin(\varphi+\theta)}{\sin(\varphi)}=\frac{x}{x+\Delta x}\frac{u\cos(\varphi)}{v\cos(\varphi+\theta)}$$ which means that: $$\frac{u^2}{v^2}=\frac{x+\Delta x}{x}\frac{\sin(\varphi+\theta)\cos(\varphi+\theta)}{\sin(\varphi)\cos(\varphi)} =\frac{x+\Delta x}{x}\frac{\sin\left(2(\varphi+\theta)\right)}{\sin(2\varphi)}\geq1$$ or, in other words, $u\geq v$ and: $$u=v\sqrt{\frac{x+\Delta x}{x}\frac{\sin\left(2(\varphi+\theta)\right)}{\sin(2\varphi)}}$$ Going back to my orginal time difference: $$\begin{align*} t_A-t_B&=\frac{2}{g}(v\sin(\varphi+\theta)-u\sin(\varphi))\\ &=\frac{2}{g}\sin(\varphi)\left(\frac{\sin(\varphi+\theta)}{\sin(\varphi)}v-u\right)\\ &=\frac{2}{g}\sin(\varphi)v\left(\frac{\sin(\varphi+\theta)}{\sin(\varphi)}-\sqrt{\frac{x+\Delta x}{x}\frac{\sin\left(2(\varphi+\theta)\right)}{\sin(2\varphi)}}\right) \end{align*}$$ But how can I judge the sign of the difference between the parenthesis? Any ideas? Thank you for your time!

Answer 1

After pondering for two days, I finally decided to answer this question, when I realized no one is going to show you (i.e. the OP) that you have already solved this yourself. In the text you have posted, you have described your attempt in detail, which hit a stonewall after some long calculations. I think, what you have done after deriving the equation (2) is unnecessary, because you could have used equations (1) and (2) to determine which projectile lands first, i.e.

$\left(a.\right)\space$ if $v\sin\left(\varphi +\theta\right)\gt u\sin\left(\varphi\right),$ then projectile $\bf{B}$ lands first,

$\left(b.\right)\space$ if $v\sin\left(\varphi +\theta\right)\lt u\sin\left(\varphi\right),$ then projectile $\bf{A}$ lands first,

$\left(c.\right)\space$ if $v\sin\left(\varphi +\theta\right) = u\sin\left(\varphi\right),$ then both projectiles land at the same time.

As far as I know, there are no other inequalities that can relay information about the landing precedence of two projectiles. If this is not the answer you required, please comment.

Answer 2

Under the given assumptions, the following are equivalent: $$\frac{\sin(\varphi+\theta)}{\sin(\varphi)}-\sqrt{\frac{x+\Delta x}{x}\frac{\sin\left(2(\varphi+\theta)\right)}{\sin(2\varphi)}}\geq 0 $$ $$1-\sqrt{\frac{x+\Delta x}{x}\cdot \frac{\sin(\varphi)^2\sin\left(2(\varphi+\theta)\right)}{\sin(\varphi+\theta)^2\sin(2\varphi)}}\geq 0 $$ $$1-\sqrt{\frac{x+\Delta x}{x}\cdot \frac{\tan(\varphi)}{\tan(\varphi+\theta)}}\geq 0 $$ $$1\geq\sqrt{\frac{x+\Delta x}{x}\cdot \frac{\tan(\varphi)}{\tan(\varphi+\theta)}} $$ $$\frac{\tan(\varphi+\theta)}{\tan(\varphi)}\geq\frac{x+\Delta x}{x}. $$

Here is another way to achieve the same result. As you stated, if a projectile in a vacuum is launched at speed $v$ at angle $\theta$, then the time of flight is $t = 2 v \sin(\theta)/g$ and the distance covered is $x=t\cdot v \cos(\theta)$. Notice that
$$ t^2 = \frac{2 v \sin(\theta)}{g} \frac{x}{v \cos(\theta)}= x \frac{2 \tan(\theta)}{g}.$$ So, if you have two projectiles and you want to know if $t_1>t_2$, you can just compute $$ t_1^2/t_2^2 = \frac{ x_1 \frac{2 \tan(\theta_1)}{g}}{ x_2 \frac{2 \tan(\theta_2)}{g}} = \frac{x_1 \tan(\theta_1)}{x_2 \tan(\theta_2)}. $$