Mathematics AA SL
Mathematics AA SL
Sample Internal Assessment
Sample Internal Assessment
11 mins Read
11 mins Read
2,060 Words
2,060 Words

To determine an analytical solution of the projectile motion of a volleyball experiencing an air resistance in terms of quadratic resistance law using analytical method & power series expansion.

Table of content


I firmly believe, mathematics has the key to unleash the quantitative aspect of any physical event and it has the ability to justify more or less any scientific event that could be observed in the human world. I, always being a mathematics lover and enthusiast of learning mathematics, IB has given me the platform to enrich my soul. Studying beyond the syllabus gives the true essence of understanding and knowledge and this has put me in front of a projectile motion. As per the curriculum of IB, I have studied about motion in a plane in Topic 2 of Physics. After studying a few books on projectile motion with resistive forces, and studying a few articles on projectiles, I came to an understanding that the projectile motion is mere an extension of the section of motion in a plane I have been taught in IB. However, the several parameters of projectile which were not discussed within the curriculum. This exploration is mainly because of determining all those parameters which affect the projectile and how does it affect the projectile motion.


The main motive of this exploration is to determine an analytical solution of the projectile motion of a volleyball experiencing an air resistance in terms of Quadratic Resistance Law using Analytical Method and Power Series Expansion.


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  • What is projectile motion

    Projectile motion is a motion of a particle when it is thrown or projected at a certain angle with the horizontal plane with a certain initial velocity. There is no acceleration along the horizontal direction; however, acceleration due to gravity acts downward throughout the motion. Thus, even if the particle is thrown with certain angle with the horizontal axis, eventually it falls back to the ground.

    What is air resistance

    Air resistance is a resistive force that any particle experiences when travels through the medium of air. When any particle moves through the layers of air, the air molecules strike with the particle and offers a resistance against the direction of motion of the particle. This resistance is known as air resistance. Air resistance increases with an increase in velocity of the motion of the particle. The expression of air resistance is shown below:


    R = mgkV2




    R = Air resistance


    m = mass of particle


    g = acceleration due to gravity


    k = proportionality constant


    V = velocity of particle

    Method of integration by parts

    The method of integration by parts is used to find the solution of any integral when the integral is in terms of product of two functions of the integrating variable. The mathematical formulation of the above-mentioned method is shown below:


    \(\displaystyle\int u.vdx\)


    \(=u\displaystyle\int vdx-\displaystyle\int \frac{du}{dx}\times\displaystyle\int vdxdx\)




    u = function of x


    v = another function of x

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  • Process of calculation

    The conventional form of certain parameters that involve in a projectile are as follows:


    \(\frac{dV}{dt}=-g\ sin\ sin\ \theta-gkV^2…(equation\ 1A)\)


    \(\frac{dθ}{dt}=\frac{g\ cos \ cos\ \theta}{V}…(equation\ 1B)\)


    \(\frac{dx}{dt}=V\ cos\ cos\ \theta...(equation\ 1C)\)


    \(\frac{dy}{dt}=V sin\ sin\ \theta …(equation\ 1D)\)


    It should be noted that the drag force is proportional to the square of the velocity.


    R = mgkV2



    V = initial velocity of point mass


    θ = angle of projection


    g = acceleration due to gravity


    m = mass of point object


    x,y = co-ordinates of the point mass in Cartesian Coordinate System


     \(k =\frac{ρcS}{2mg}=\frac{1}{V^2_{term}}=constant\)


    k = proportionality constant


    ρ = density of air


    c = coefficient of drag


    S = cross sectional area of the object


    V = terminal velocity of point mass


    The above parameters are shown in the schematic diagram of the projectile motion in Fig 1.

    Figure 1 - Schematic Representation Of Parameters Of Projectile Motion

    Equation 1A, 1B, 1C, and 1D depends on the velocity of the projectile which further depends on the angle between the slope of the projectile motion at any instant.


    \(V(\theta)=\frac{V_0\ cos\ \theta_0}{cos\ \theta\sqrt{1+V_0^2\ cos^2\ \theta_0(f(\theta_0)-f(\theta))}}\)


    \(f(\theta)=\frac{sin\ \theta}{cos^2\ \theta}+In\ tan\bigg(\frac{\theta}{2}+\frac{\pi}{4}\bigg)\)          (equation - 2)


    \(t=t_0-\frac{1}{g}\displaystyle \int_{\theta_0}^{\theta}\frac{V}{cos\ \theta}d\theta\)


    \(x=x_0-\frac{1}{g}\displaystyle \int_{\theta_0}^{\theta}V^2d\theta\)


    \(y=y_0-\frac{1}{g}\displaystyle \int_{\theta_0}^{\theta}V^2\ tan\ \theta\ d\theta\)          (equation - 3)



    V= initial velocity


    θ0 = initial slope of trajectory


    t= initial value of time


    x= initial value of abscissa of the location of point mass


    y= initial value of ordinate of the location of point mass


    As the integral that is shown in the RHS of equation (3) cannot be solved using elementary functions, in order to find the values of variables t, x, and y, analytical equations are derived in the form of a series of functions. Using integration by parts of the quadratures, the integral has been represented within a finite range of \([\theta_0,\theta]\) as a power series of functions.

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  • Construction of series of function

    Let, variable u = Vcos cos θ. u is the horizontal component of velocity.


    As it is well known [13], this auxiliary variable satisfies the following differential equation under the quadratic resistance law:




    The resistive force is represented by R


    In this exploration, the value of R < 1 has been considered which defines that the motion of the projectile is dominating over the resistive force.


    From equation 2, 3, and 4:.


    \(g\left\{\begin{matrix}t\\x\\y\\\end{matrix}\right\} = - \displaystyle\int\limits_{\theta _0}^{\theta} \left\{\begin{matrix}u\frac{1}{cos^2 \theta}\\u^2\frac1{cos^2 \theta}\\u^2\frac{sin \theta}{cos^2 \theta}\\\end{matrix}\right\}d \theta\)


    Equation (5)


    Firstly, the dependencies are obtained:


    t = t(θ), x = x(θ)


    The integral is solved on the (i - 1)th step of integration by parts:




    Then i - th step is as follows:


    \(I_{i-1}=\frac{1}{k}\displaystyle\int u^3\frac{dZ_{i-1}(u)}{du}\cdot\frac{X_{i-1}(\theta)}{cos^3\ \theta}d\theta\)


    \(=\displaystyle\int Z_i(u)\frac{dX_i(\theta)}{d\theta}d\theta=Z_i(u)X_i(\theta)-kI_i\)






    Equation (6)


    Hence, t and x obtained are expressed as functional series:


    \(g\left\{\begin{matrix}t\\x\\\end{matrix}\right\} = \frac1k \sum\limits_{i=1}^∞ (-k)^i\left\{\begin{matrix}F_i(u) \cdot \Phi_i(\theta)\\H_i(u) \cdot\Phi_i(\theta)\\\end{matrix}\right\}\)


    Equation (7)


    Here the functions Fi(u), Hi(u),Φi(u) are related by the recurrence relations as shown in equation (6). Here, the functions that are generated are as follows:


    \(F_1(u)=u,\ H_1(u)=u^2,\ Φ_1(\theta)\)


    \(= \displaystyle\int\frac{d\theta}{cos^2\theta}=tg\theta\)


    \(F_i(u)=(2i-3)!!u^{2i-1},\ H_i(u)\)




    Equation (8)


    The general structures of the following functions expressed by  are deduced using the formulae present in the integration table:


    \(Φ_{2n-1}(\theta)=sin\ \theta\cdot\sum\limits^{2n-2}_{m=0}A_{2n-1,m}\cdot\frac{1}{(cos\ \theta)^{2m+1}},\)


    \(Φ_{2n}(\theta)=\sum\limits^{2n-2}_{m=0}A_{2n,m}\cdot\frac{1}{(cos\ \theta)^{2m+3}},n=1,2,3...\)


    Ai,m = constant coefficient


    The function Φ2n, Φ2n-1 has been substituted in equation (6) and the coefficients of equal power are compared:


    A1,0 = 1








    The coefficients A2n+1,m are reduced using Binomial expansion:




    These relations do not determine the coefficients A2n+1,0, but they are defined by the general formula for A2n+1,m.


    \(\displaystyle\int u^3\cdot\frac{dZ_{i-1}}{du}\cdot\frac{d\theta}{cos^3\ \theta}\)




    \(I_{i-1}=\frac{1}{k}\displaystyle\int \frac{dH_{i-1}(u)}{d\theta}\cdot\Psi_{i-1}(\theta)d\theta\)


    Then i – th step is performed as follows:


    \(I_{i-1}=\displaystyle\int u^3\frac{dH_{i-1}(u)}{du}\cdot\frac{\Psi_{i-1}(\theta)}{cos^3\ \theta}d\theta\)


    \(=\displaystyle\int u^3 \frac{dH_{i-1}(u)}{du}\)


    \(\bigg(\frac{\alpha_{i-1}}{cos^3 \theta}+\frac{\Psi_{i-1}(\theta)-\alpha_{i-1}}{cos^3\ \theta}\bigg)d\theta\)


    \(=\frac{1}{k}H_{i-1}(u)\alpha_{i-1}+\displaystyle\int H_i(u)\frac{d\Psi_i(\theta)}{d\theta}\)










    Equation (9)


    where αi-1 compensating constants to be determined. As a result, the y coordinate a series of functions of the form has been obtained:




    Equation (10)


    where the functions Hi(u) and Ψi(u) are connected by recurrence relations (9).


    \(H_1(u)=u^2,Ψ_1(\theta)=\displaystyle\int\frac{sin\ \theta\ d\ \theta}{cos^3\ \theta}=\frac{1}{2\ cos^2\theta}\)


    Functions Hi(u) have been obtained (8). A common structure of functions Ψi(θ) is now been obtained:


    \(\left.\begin{matrix}𝛙_1(\theta) = \frac1{2cos^2 \theta} = \\ \frac1{cos^2 \theta} \cdot 𝛙_1^*(\theta)𝛙_{2n}(\theta) = \\ \\ sin(\theta) \cdot \sum\limits_{m=0}^{2n-2}B_{2n,m} \cdot\frac1{(cos \theta)^{2m\ +\ 4}} \\ = \frac1{(cos \theta)^{4n}} \cdot 𝛙^*_{2n}(\theta)\\ \\𝛙_{2n+1}(\theta) = \sum\limits_{m=0}^{2n-2}B_{2n+1,m}\ \cdot \ \frac1{(cos \theta)^{4n + 2}} \\ = \frac1{(cos \theta)^{4n + 2}}\ \cdot \ 𝛙^*_{2n+1}(\theta)\\\end{matrix}\right\}\)

    n = 1.2.3....


    Equation (11)


    Here Bi,m are the constant coefficients. Introduced functions Ψi*(θ)trigonometric polynomials of order (2i - 6) for odd values i, and (2i - 3) for even values of i. Substituting the functions Ψ2n+1, Ψ2n in (9) and equating the coefficients of equal powers of cos  , we obtain:


    \(\alpha_{2n}=0\ ;\ B_{2,0}=\frac{1}{8};\ B_{2n+1,m}\)




    Equation (12)




    Equation (13)


    \(B_{2n+2,m}=\frac{1}{4+2m}\cdot B_{2n+1,m-2}+\)




    \(\cdot B_{2n+1,j}\)






    Equation (14)


    \(\left.\begin{matrix}t=\frac{1}{gkV}\cdot\sum\limits^\infty_{i=1}(2i-3)!! \\ (-r)^i \ \cdot\Phi_i^*(\theta)\bigg|^\theta_{\theta_0}\\ \\x=\frac{cos\theta}{gk}\cdot\sum\limits^\infty_{i=1}(2i-2)!! \\ (-r)^i\ \cdot\Phi^*_i(\theta)\bigg|^\theta_{\theta_0},\\ \\y=\frac{1}{gk}\sum\limits^\infty_{i=1}(2i-2)!!. \\ (-r)^i(\Psi^*_i(\theta)-\alpha_icos^{2i}\theta)\bigg|^\theta_{\theta_0}\cdot\\\end{matrix}\right\}\)


    Equation (15)


    In this exploration, we obtained an analytical solution of the problem of the motion of heavy projectile with a quadratic resistance at an arbitrary angle of throwing. The solution is obtained using the classical method of calculating integrals by parts and is written in the form of power series. It is shown that only six members of the series provide high accuracy of construction of the projectile trajectory. For comparison, we note that in [6] the order of approximation of the projectile trajectory is 7, in [7] mainly 12 terms of the series were used for an acceptable approximation of the trajectory, and in [8] the 60-th order of approximation was used. Thus, the proposed approach which based on the use of analytical formulae makes it possible to simplify significantly a qualitative analysis of the motion of a projectile with the quadratic drag force.


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    • Integration by Parts. Accessed 9 Mar. 2021.
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    • Free Fall and Air Resistance. Accessed 9 Mar. 2021.
  • Nail IB Video
    Patrick Jones

    1.2M+ YouTube subscribers, 30+ years' experience, Ivy League visiting faculty!

    Video Course

  • Nail IB Video
    Patrick Jones

    1.2M+ YouTube subscribers, 30+ years' experience, Ivy League visiting faculty!

    Video Course