Question: A ball is thrown at speed $v$ from zero height on level ground. At what angle should it be thrown so that the distance traveled through the air is maximum.

As I solve the displacement in $x$ and $y$ component, I got $$x=v \cos(\theta)t$$ and $$y=v \sin(\theta)+1/2gt^2.$$

When the ball reach maximum height, the trajectory length will be integrate $$\sqrt{(dx/dt)^2+(dx/dt)^2}.$$ Can anyone explain to me the last part, why do we integrate that value?

  • $\begingroup$ Shouldn't there be a "t" attached to the velocity parts of the x and y displacement formula that you gave? Also the integral needs bounds and a variable that you integrate over $\endgroup$ – J. Shupperd Oct 25 '16 at 19:14
  • $\begingroup$ Sorry, I missed that but I corrected it $\endgroup$ – Tammy Chong Oct 25 '16 at 19:19
  • $\begingroup$ Try here mathhelpboards.com/calculus-10/maximizing-arc-length-2167.html . The method is simple - find $y(x)$ for arbitrary $\theta$, find the length of thr trajectory for this arbitrary $\theta$ (~ $\int \sqrt{1+y'^2}dx$) and then find the maxinal value of this (length) function. $\endgroup$ – Alexander Oct 25 '16 at 20:12

Unlike the (easier) problem where you are asked to maximize the $x$ distance traveled before it hits the ground, here you were asked to maximize the path length, that is, the total length of the curve (in x-y space) folowed by the ball.

What is the length traveled in a tiny time interval $\delta t$?

Well, in that tiny interval, you can consider that the particle is moving in nearly a straight line, and that its velocity does not change. So during $\delta t$ it will move horizontally by an amount $\delta x = \frac{dx}{dt} \delta t$ and it will move vertically by $\delta y = \frac{dy}{dt} \delta t$. Then by the pythagorean theorem, the overall length traveled is $$ \delta s = \sqrt{(\delta x)^2+(\delta y)^2}= \sqrt{\left(\frac{dx}{dt}\delta t \right)^2+\left(\frac{dy}{dt}\delta t \right)^2 } =\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 } \delta t $$ and since the ball hits the ground at time $T = 2\frac{v_y}{g} = 2\frac {v\sin \theta}{g}$ the total curve length is $$ L(\theta) = \int_{t=0}^{2\frac {v\sin \theta}{g}}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2 } dt $$ If we insert the expressions for the derivatives of $x(t)$ and $y(t)$ this becomes $$ L(\theta) = \int_{t=0}^{2\frac {v\sin \theta}{g}}\sqrt{v^2\cos^2\theta+\left(v\sin\theta-gt\right)^2 } dt $$ which simplifies to $$ L(\theta) = v \int_{t=0}^{2\frac {v\sin \theta}{g}}\sqrt{v^2\cos^2\theta+v^2\sin^2\theta -2vgt\sin\theta +g^2t^2} dt $$ If we replace $\sin \theta = u$ this becomes $$ L(u) = v \int_{t=0}^{2\frac {v u}{g}}\sqrt{g^2t^2-2gtuv+v^2} dt = \frac{v^2}{2g}\left( 2u + (1-u^2) \log\left( \frac{1+u}{1-u}\right)\right) $$ And then $$ \frac{dL(u)}{du} = \frac{v^2}{2g} \left(4-2u\log\left(\frac{1+u}{1-u}\right)\right)$$ If you numerically solve $$ 4-2u\log\left(\frac{1+u}{1-u}\right)=0$$ you get $u = 0.83355656$ which represents an angle of about $\theta = 56.4658^\circ$ and a total arc length of $2.39936 \frac{v^2}{g}$, as compared to only $2.29559 \frac{v^2}{g}$ for $\theta = 45^\circ$.

  • $\begingroup$ I do not understand the last two step, why dL(u)/du? and how to get the equation where i can solve it numerically? $\endgroup$ – Tammy Chong Oct 26 '16 at 7:53
  • $\begingroup$ If i solve the integral using Mathematica instead of substitution, how do i proceed from the answer of the integral? I am stuck $\endgroup$ – Tammy Chong Oct 26 '16 at 7:55

Can anyone explain to me the last part, why do we integrate that value?

In an infinitesimal amount of time $dt$ the projectile travels $dx$ in the $x$-direction and $dy$ in the $y$-direction:


The total length traveled is $dl$ and found with Pythagoras:


With $dx=\frac{dx}{dt}dt$ and $dy=\frac{dy}{dt}dt$:



$$\implies l=\int_0^t\sqrt{\Big(\frac{dx}{dt}\Big)^2+\Big(\frac{dy}{dt}\Big)^2}dt$$

Where $t$ is the total flight time.

But to find the optimum angle you don't need to compute that integral. Instead, compute:

$$\frac{\partial l}{\partial \theta},$$

from $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

Optimum theory tells us that for:

$$\frac{\partial l}{\partial \theta}=0,$$

the trajectory length will be optimal.

  • $\begingroup$ I understood the first part and can proceed to the next steps, but i have no idea about the optimum theory. Is this method simpler? $\endgroup$ – Tammy Chong Oct 26 '16 at 6:50
  • $\begingroup$ It's the same as 'Mark's' solution. You're looking for the maximum of $l(\theta)$. Well, for a function $f(x)$ we find the optimum $x$ for $\frac{d}{dx}f(x)=0$. So first you find $l(\theta)$, then solve $\frac{d}{d\theta}l(\theta)=0$. It's really the only way. $\endgroup$ – Gert Oct 26 '16 at 14:01

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