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Recently I was wondering about what will happen to the particle when subjected various elevation angle of projectile motion if we account air resistance. I want to know what the angle of elevation that can make the largest area that projectile can cover.

First of all, to make things easier, I will take case on ideal projectile motion, and my method of solving these problem. I also worked with Wolfram Mathematica in this problem

Ideal Condition (maximum area covered)

To me, my method to find the maximum area covered is quite debatable, but somehow, my method is working. My target is try to make a Area function for elevation angle $\theta$

As we obviously know that:

$x\left(t\right)=x_0+v_i\cos{\left(\theta\right)}t$

$y\left(t\right)=y_0+v_i\sin{\left(\theta\right)}t-\frac{1}{2}gt^2$

Assuming the motion start in origin (0,0), then we can drop the term $y_0$ and $x_0$. Combining the same term for $t$, I can get the equation of trajectory:

$y\left(x\right)=\tan{\left(\theta\right)}x-\frac{g}{2{v_i}^2\cos^2{\left(\theta\right)}}x^2$

Defining the the Area function:

$A\left(t\right)=x\left(t\right)y\left(t\right)$

$A\left(t\right)=\frac{v_it^2}{2}\left(v_i\sin{\left(2\theta\right)}-g\cos{\left(\theta\right)t}\right)$

Then, derive with respect to $t$ and set equal to zero.

$\frac{d}{dt}\left(A\left(t\right)\right)=0$

$\frac{d}{dt}\left(\frac{v_it^2}{2}\left(v_i\sin{\left(2\theta\right)}-g\cos{\left(\theta\right)t}\right)\right)=0$

$t=\frac{2\sin{\left(2\theta\right)}}{3g\cos{\left(\theta\right)}}$

This is the time that the trajectory must achieve to make largest area covered. Return this time value back equation $A(t)$, I can get the Area equation for elevation angle, $A(\theta)$.

$A\left(\theta\right)=\frac{2\ v^2\ {\sec{\left(\theta\right)}}^2{\sin{\left(\theta\right)}}^3}{27\ g^2}$

This is the function that i want, now derive this equation again with respect to $\theta$ and set it equal again to zero. Finally solve for $\theta$. Let use Mathematica to get the elevation angle.

enter image description here

As it can be seen, there are 6 solution of the angle, we can just take one of it in which lies in quadrant I. Which is $\theta=\frac{\pi}{3}$. I can verify this answer by making a simple program in GeoGebra. You can click it here to verify your own. As it can see the largest area that cover is when the elevation angle is $\theta=\frac{\pi}{3}$

General Condition (maximum area covered)

From the previous method, is seems like its quite good approximation to get the right elevation angle for largest area covered. To summarize form previous method is as follow:

  1. Find the equation of motion in x & y plane with respect for time
  2. Combining the two equation previously, and get the equation of trajectory $y\left(x\right)$
  3. Defining Area-time function : $A\left(t\right)=x\left(t\right)y\left(t\right)$
  4. Solving $t$ from $\frac{d}{dt}\left(A\left(t\right)\right)=0$
  5. Return the value $t$ back to $A\left(t\right)$ and get the new function as $A\left(\theta\right)$
  6. Solving $\theta$ from $\frac{d}{dt}\left(A\left(\theta\right)\right)=0$

For the drag force, I use the linear drag force. $F_d=-bv$ and the direction of force is always opposite to the direction of motion. By doing some work in Newtonian mechanics, I can get the equation of motion versus time.

$x\left(t\right)=x_0+\frac{mv\cos{\left(\theta\right)}}{b}\left(1-e^{-\frac{b}{m}t}\right)$

$y\left(t\right)=y_0+\frac{m}{b}\left(\frac{mg}{b}+v\sin{\left(\theta\right)}\right)\left(1-e^{-\frac{b}{m}t}\right)-\frac{mg}{b}t$

With $b$ is some constant drag coefficient, and $m$ is particle mass. For the sake of simplicity, the particle move from the origin (0,0). This will make $y_0$ and $x_0$ disappeared.

To find the area of the area function against time, multiplying the two combinations of equations of motion ($x(t)$ and $y(t)$). Multiplying these two functions will produce a function of area with respect to time $A\left(t\right)$. Derive $\frac{d}{dt}\left(A\left(t\right)\right)=0$, and find a solution for $t$. I will see if there is a solution of t from this equation. The Mathematica syntax for this problem is as follows:

enter image description here

This is the real question begin. Mathematica cannot solve this equation. I'm skeptical that from the beginning my method was false, or it really doesn't have any method to solve the equation for elevation angle that makes the largest area covered from projectile motion when accounting the air resistance?

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1 Answer 1

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I don't think that your feature to optimized the starting angle is correct ?

enter image description here

if you look at this graphics, the red one is the starting position , if you change the angle $~\theta~$ , to obtain the same end position ,you have to change also the starting velocity $~v_0$. your feature just optimized the angle $~\theta~$ ?


$$y(x)=\frac 12\,{\frac {g{x}^{2}}{m{{ v_0}}^{2} \left( \cos \left( \theta \right) \right) ^{2}}}+{\frac {x\sin \left( \theta \right) }{\cos \left( \theta \right) }} $$ $$A=\int_0^{x_\rm{red}}\,y(x)\,dx=A(\theta~,v_0)$$

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  • $\begingroup$ what if the initial velocity is determined, and I only want to change the angle to make the most large area reached. Im not making the trajectory to make the same starting and ending position $\endgroup$ Nov 14, 2022 at 1:37
  • $\begingroup$ It doesn’t make sense for me if the starting and ending position are not equal, for all trajectories? $\endgroup$
    – Eli
    Nov 14, 2022 at 7:14

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