Physics Friday 95

Consider a projectile launched from the ground at an initial velocity v0 at an angle θ above the horizontal, with negligible air resistance. The projectile will thus follow a parabolic trajectory. What is the angle θ such that the total distance the projectile travels in the air is maximized? Note, this is the length of the parabolic trajectory, not the distance along the ground to the impact point, and so differs from the h=0 case of Physics Friday 24.


The only acceleration is due to gravity; so our basic kinematic equations tell us that x=(v_0\cos\theta)t and y=(v_0\cos\theta)t-\frac{1}{2}gt^2.
Solving for t as a function of x and substituting, we get
t=\frac{x}{v_0\cos\theta}, and thus
y=(v_0\cos\theta)\frac{x}{v_0\cos\theta}-\frac{1}{2}g\left(\frac{x}{v_0\cos\theta}\right)^2=x\tan\theta-\frac{g}{2v_0^2\cos^2\theta}x^2.
The trajectory ranges between the two x values where y=0, which are x=0 and x=\frac{2v_0^2\sin\theta\cos\theta}{g}=\frac{v_0^2\sin(2\theta)}{g}.

Thus, we find the length of the parabola via the arc length formula
\begin{array}{rcl}L&=&\int_{0}^{v_0^2\sin(2\theta)/g}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\\&=&\int_{0}^{v_0^2\sin(2\theta)/g}\sqrt{1+\left(\tan\theta-\frac{gx}{v_0^2\cos^2\theta}\right)^2}\,dx\end{array}
Making the substitution u=\tan\theta-\frac{g}{v_0^2\cos^2\theta}x, we get
\begin{array}{rcl}L&=&\int_{0}^{v_0^2\sin(2\theta)/g}\sqrt{1+\left(\tan\theta-\frac{gx}{v_0^2\cos^2\theta}\right)^2}\,dx\\&=&\int_{\tan\theta}^{-\tan\theta}\sqrt{1+u^2}\left(-\frac{v_0^2\cos^2\theta}{g}\right)\,du\\&=&\frac{v_0^2\cos^2\theta}{g}\int_{-\tan\theta}^{\tan\theta}\sqrt{1+u^2}\,du\\&=&\frac{2v_0^2\cos^2\theta}{g}\int_{0}^{\tan\theta}\sqrt{1+u^2}\,du\end{array}.
(the last step is due to symmetry, as the integrand is an even function; see here.)
Now, by consulting a table of integrals (or via trigonometric substitution), we find that
\int\sqrt{1+x^2}\,dx=\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\ln\left(x+\sqrt{1+x^2}\right)+C,
and thus our length becomes
\begin{array}{rcl}L&=&\frac{2v_0^2\cos^2\theta}{g}\int_{0}^{\tan\theta}\sqrt{1+u^2}\,du\\&=&\frac{2v_0^2\cos^2\theta}{g}\left[\frac{u}{2}\sqrt{1+u^2}+\frac{1}{2}\ln\left(u+\sqrt{1+u^2}\right)\right]_{0}^{tan\theta}\\&=&\frac{2v_0^2\cos^2\theta}{g}\left(\frac{\tan\theta}{2}\sqrt{1+\tan^2\theta}+\frac{1}{2}\ln\left(\tan\theta+\sqrt{1+\tan^2\theta}\right)\right)\\&=&\frac{v_0^2\cos^2\theta}{g}\left(\tan\theta\sec\theta+\ln\left(\tan\theta+\sec\theta\right)\right)\\L&=&\frac{v_0^2}{g}\left(sin\theta+\cos^2\theta\ln\left(\frac{1+\sin\theta}{\cos\theta}\right)\right)\end{array}.

To determine the maximum, we find \frac{dL}{d\theta} and set it equal to zero.
Now, taking the derivative, we get
\begin{array}{rcl}\frac{dL}{d\theta}&=&\frac{d}{d\theta}\frac{v_0^2}{g}\left(sin\theta+\cos^2\theta\ln\left(\frac{1+\sin\theta}{\cos\theta}\right)\right)\\&=&\frac{v_0^2}{g}\left[\cos\theta+\frac{(cos^2\theta+(1+\sin\theta)\sin\theta)\cos\theta}{1+\sin\theta}\right.\\&&\left.-2\sin\theta\cos\theta\ln\left(\frac{1+\sin\theta}{\cos\theta}\right)\right]\\&=&\frac{2v_0^2}{g}\cos\theta\left[1-\sin\theta\ln\left(\frac{1+\sin\theta}{\cos\theta}\right)\right]\end{array}.

Since we want 0<θ<π/2, the above will be zero when
1-\sin\theta\ln\left(\frac{1+\sin\theta}{\cos\theta}\right)={0}
\sin\theta\ln\left(\frac{1+\sin\theta}{\cos\theta}\right)=1.
(Note that this is an angle independent of our initial velocity.) However, the above equation cannot be solved analytically; a numerical solution gives θ≈0.98551≈56.466°.

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