In a projectile motion, consider the projectile to be launched from $(0,0)$ as traced to be on cartesian axes. Then the trajectory is given by
$$y=x \tan \theta \big(1-x/R)$$
where $\theta$ is measured from $x$ axis , and gravity acts along $y$ axis, $R$ stands for the maximum range which may vary with the initial velocity which has a magnitude of $v$.
Using arc length formula of a general curve $\Gamma$,
$$\ell(\Gamma)=\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx$$
we get the distance traversed as an integral of form $D=\sqrt{ax^2+bx+c}$.
An interactive model can be found here.
I would like to know at what angle $\theta$ the distance travelled will be maximized, given a fixed $v$.
I tried doing
$$\frac{d}{d \theta}D=0$$ but it gave me an unknown $\frac{dx}{d \theta}$, I am unable to approach further, so I created a graph here which still doesn't helps to get a fixed value.
Best Answer
The projectile equations are:
$$x=v\,\cos(\varphi)\,t\tag 1$$ $$y=v\,\sin(\varphi)\,t-\frac{1}{2}\,g\,t^2\tag 2$$
the arc length is:
$$L=\int_0^{t_f} \sqrt{\dot{x}^2+\dot{y}^2}\,dt$$
with $y=0~,\Rightarrow $ $t_f=\frac{2\,v\sin(\varphi)}{g}$
thus:
$$L=L(v~,\varphi)\tag 3$$
and the max arc length condition :
$$\frac{\partial L}{\partial \varphi}=\\\frac{v^2\,\cos(\varphi)}{g}\,\underbrace{(2+\ln \left( -\sin \left( \varphi \right) +1 \right) \sin \left( \varphi \right) -\ln \left( \sin \left( \varphi \right) +1 \right) \sin \left( \varphi \right) )}_{f(\varphi)}=0$$
$f(\varphi)=0~\Rightarrow~ \varphi=56.465^\circ$
Remarks :
I used Maple to do the calculations.