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I know that $45°$ gives the longest range, and I feel as though $45°$ might also be the angle for the longest path of the object, but I'm not sure. How could one find what this angle is? It may also be that this angle relies on the initial velocity, so it would not be a constant.

To be clear what I mean by path, imagine that the object releases string into the air as it travels. The length of the path would be the length of that string.

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    $\begingroup$ The arc length (or path length) is expressed mathematically as $S=\int \sqrt{1+(y'(x))^2}dx$. Since writing $y(x)$ is very simple (its a square function of $x$ which depends also on the angle), you can express the the path length as an analytic function of the angle. You can then take the derivative with respect to the the angle to find the maxima. Since, the dependence is expressed using trigonometric functions solving for the maxima might not be feasible analytically, and it is also possible to have multiple solutions. $\endgroup$
    – Yair M
    Oct 24, 2017 at 18:25
  • $\begingroup$ Don't assume 45° gives the longest range (unless given in a problem statement). See Dj_Algebra's answer. $\endgroup$ Oct 24, 2017 at 18:57
  • $\begingroup$ 45 gives the longest range for any initial velocity $\endgroup$
    – volcanrb
    Oct 24, 2017 at 19:21
  • $\begingroup$ I just did it. If you pursue the calculation I suggested, you'll see that as long as the projectile is thrown from a level plain, the longest distance corresponds to an angle of 90 degrees. $\endgroup$
    – Yair M
    Oct 24, 2017 at 19:28
  • $\begingroup$ @YairM You got the wrong root in the derivative. The longest path is for $\psi = 56.4658°$ with $$ \frac{g \ell}{v^2} = 1.1998$$ $\endgroup$ Oct 24, 2017 at 19:29

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There is likely a neat trick to solving this, but here is a first crude stab at a solution:

Let the projectile start from the origin at speed $v_0=1$ and angle $\theta$, tracing out the curve $$x(t)=\cos(\theta) t$$ $$y(t)=\sin(\theta)t-gt^2/2.$$ It will land after time $t_{max}=2\sin(\theta)/g$. The length of the trajectory is $$L=\int_0^{t_{max}} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt$$ $$=\int_0^{t_{max}} \sqrt{\cos^2(\theta)+\sin^2(\theta) + g^2t^2 - 2\sin(\theta) g t} dt$$ $$=\int_0^{t_{max}} \sqrt{1 + g^2t^2 - 2\sin(\theta) g t} dt.$$ We can reparametrize $u=gt$, $$L=(1/g)\int_0^{2\sin(\theta)} \sqrt{1 + u^2 - 2\sin(\theta) u} du.$$

We want to find $dL/d\theta=0$. By symmetry we know $\theta=\pi/2$ must be an extremum. We can either try to evaluate the integral and then find a root, or take the derivative under the integral sign (remembering that one of the boundaries also depends on $\theta$). In either case I get a messy expression in my symbolic calculator I suspect actually simplifies nicely if one massages it in the right way.

In any case, plotting $L(\theta)$ shows that it has a maximum just below $\theta=1$ (I get an angle of $56.465^\circ$). The vertical $\theta=\pi/2$ trajectory is a local minimum: adding a bit of horizontal velocity increases the length. Projectile distance as function of $\theta$ for $g=9.82, v_0=1$

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  • $\begingroup$ This is the correct answer. The dimensionless length $\frac{g\,L}{v^2}$ should peak out at about $\approx 1.2$ $\endgroup$ Oct 24, 2017 at 19:48
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Make an equation of distance traveled by the projectile in terms of initial angle of projection and also in terms of initial velocity. For taking out maximum distance equate the derivative of the above equation w.r.t dist. to 0.This will give you the angle for max range. Also note that angle 45 degree gives max range only for projectiles on horizontal surface, but this angle might vary if projectile is thrown from inclined plane, so always go by derivative method.

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As a note to add to the answer by Andrew the analytical solution is: $$\frac{dL}{d\theta}=\sin \theta\cos \theta\left( 1+ \frac{2}{\sin\theta}+2\ln(1-\sin \theta) \right)=0$$

Which gives the correct roots: enter image description here

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enter image description here

It does, as you mentioned, rely on the initial velocity, but results will stay as long as all launches have the same velocity. But I encourage you to refer to the attached photograph. Notice each path's footnote data. The highest angle recorded (other than 90), has the highest time, (T-1.93s). Follow this to higher angles and get 90 degrees as an answer (t-2.00s).

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  • $\begingroup$ This does not address the question asked. $\endgroup$
    – Kyle Kanos
    Oct 25, 2017 at 10:02

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