0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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 $\endgroup$ Oct 25, 2016 at 19:14
  • $\begingroup$ Sorry, I missed that but I corrected it $\endgroup$ Oct 25, 2016 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, 2016 at 20:12

2 Answers 2

3
$\begingroup$

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 $\theta = 45^\circ$.

$\endgroup$
2
  • $\begingroup$ I do not understand the last two step, why dL(u)/du? and how to get the equation where i can solve it numerically? $\endgroup$ Oct 26, 2016 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 $\endgroup$ Oct 26, 2016 at 7:55
1
$\begingroup$

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:

Trajectory

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

$$dl^2=dx^2+dy^2$$

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

$$dl^2=\Big[\Big(\frac{dx}{dt}\Big)^2+\Big(\frac{dy}{dt}\Big)^2\Big]dt^2$$

$$dl=\sqrt{\Big(\frac{dx}{dt}\Big)^2+\Big(\frac{dy}{dt}\Big)^2}dt$$

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

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.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Oct 26, 2016 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, 2016 at 14:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.