maximum range of projectile launched from elevation

Question

I am trying to find the maximum range of projectile from an elevation. I found the answer in this question, but I have two questions:

1. Why does $y$ need to be $0?$
2. Why do they differentiate with respect to $\theta,$ and what is the meaning of this?

Answer 1

1. Why does $y$ need to be $0?$

Because the projectile's range $R$ is its horizontal distance travelled, which is precisely its $x$-coordinate at its landing point, that is, when there is no elevation, that is, when $y=0.$

2. Why do they differentiate with respect to $\theta,$ and what is the meaning of this?

The projectile's range $R$ varies according to its launch angle $\theta,$ so is maximised when $\frac{\mathrm dR}{\mathrm d\theta}$ equals zero. (Consider the graph of $R$ against $\theta;$ $R$ attains its maximum value when its gradient $\frac{\mathrm dR}{\mathrm d\theta}$ equals zero.)

P.S. I am the author of the cited Question; even though I accepted Patrick's Answer, please treat my self-Answer there as the natural continuation of the work that I'd started in the Question.

Answer 2

Let projectile to be shot at angle $\theta$. Then $x=ut\cos\theta$, $y=H+ut\sin\theta - \frac{gt^2}{2}$. From first equation $t=\frac{x}{u\cos \theta}$, then $y=H+x \tan\theta-\frac{gx^2}{2u^2\cos^2\theta}=H+x \tan\theta-\frac{gx^2}{2u^2}(1+\tan^2\theta)$.

Range of projectile at some $\theta$ is $x$ at which $y=0$. Then range $R$ is defined by $H+R\tan\theta-\frac{gR^2}{2u^2}(1+\tan^2\theta)=0$. Let mark $\tan\theta=p$, then $H+Rp-\frac{gR^2}{2u^2}(1+p^2)=0$.

Let consider this equation with respect to $p$ and rewrite it as $$\frac{gR^2}{2u^2}p^2-Rp+\left(\frac{gR^2}{2u^2}-H\right)=0.$$ Solution for $p$ will exist if $$D=R^2-4\frac{gR^2}{2u^2}\left(\frac{gR^2}{2u^2}-H\right)\geq 0$$

$$R^2>0\Rightarrow 1-4\frac{g}{2u^2}\left(\frac{gR^2}{2u^2}-H\right) \geq 0\Rightarrow \frac{gR^2}{2u^2}-H \leq \frac{u^2}{2g} \Rightarrow R^2\leq \frac{u^4}{g^2}+\frac{2u^2}{g}H$$ $$R\leq \sqrt{\frac{u^4}{g^2}+\frac{2u^2}{g}H}=\frac{u}{g}\sqrt{u^2+2gH}$$