Effect of quadratic air resistance on the trajectory of a Shot Put (Projectile)

Mathematics is the tool which quantifies any physical, chemical or biological event. Not only condensed within the boundaries of the domain of science and technology, but also it has enriched other domains of knowledge such as, management, business, anthropology and many more with its applications. The scope that particular subject provides, with such diverse domain of application which leads to the significant opportunity to interpret not only scientific events, but also commerce and humanity topics, has immensely intensified my interest towards this subject. For example, physics and mathematics are often termed as either sides of a coin.

 

Due to global pandemic, the occurrence of physical events that usually ignites the curiosity in my mind has decreased followed by little difficulty in identifying the applications of mathematics in the physical world for the mathematics internal assessment. But being an inquirer, I tried to see the world on the screen of television. While watching several athletic events as well as the most famous sport, ‘Football’, I found that wind has a significant role to play in these sporting events. However, it could bring in partiality in the games where the participants take part in the game one after another. For example, as football is a game where both the teams play together, so any external event such as rain, wind or snow would equally impact both the teams, but in games like javelin, shot put, archery and others, the participants play in turns (one after another). As a result, external agents such as wind, rain and other factors could be present as well as absent for during the turns of different players. Hence, the effect of wind in these sports should be understood in order to establish an error free judgement.

 

To explore the effect of wind on these sports, I have taken a few tutorials in Physics to learn the concept of projectile motion, few applications of calculus, such as local maximum and minimum and also read a few journals on the effect of air resistance in projectile. After doing bit of a research, I found a journal on the effect of linear air resistance on motion of javelin; however, I could neither find any journal on effect of quadratic air resistance on projectile motion nor did I find the effect of air resistance on shot put. As the mass of shot put is significantly more than that of a javelin, how does the mass affect the trajectory of projectile in presence of air resistance? Along with this, a few more questions came into my mind for which I could not find any answer. To decipher the answers of those questions, I have chosen this as a topic for my Mathematics Internal Assessment.

The main motive of this exploration is to determine the effect of quadratic air resistance (wind current) on the trajectory of a shot put (projectile) and compare that with respect to a controlled state (absence of air resistance).

It is a global played sporting event in which each participant has to throw a spherical ball of metal of certain mass from the marked throwing position in an open field. The athlete gets an area of the field which is marked by lines in both the sides. These two lines are made keeping an angle 34.92° in between. The player should throw the shot put to avoid foul. The participant who throws the shot put to the maximum distance from the throwing position is announced to be the winner. For men, the weight of the shot put is 7.26 kg with a diameter of 110 mm to 130 mm. However, for female, the weight of the shot put is 4.00 kg with a diameter of 95 mm to 110 mm. The shot put be thrown with one arm and the position of the arm should be above the shoulder.

 

Projectile motion defines the trajectory as well as several parametric quantities of the motion of a small mass or projectile in a plane (two dimension). In general, two axes of the projectile motion are considered as the horizontal axis with respect to the direction of the motion of the projectile and the vertical axis of the Earth. The object is thrown or projected in the open space with some angle with respect to the horizontal axis of the Earth. As a result, it executes some displacement in the horizontal axis corresponding to the direction of the throw as well as in the vertical axis. Hence, it is also known as motion in a plane or two-dimensional motion. The parametric quantities of a projectile motion are as follows:

• Displacement: Total or net distance traversed by the particle in a particular direction. As the motion is being executed in both the axes, the particle executes two displacements (one for horizontal direction and another for vertical direction) at a same time.
• Velocity: It is the displacement done by the particle or projectile in unit time in a particular direction. Similar to displacement, a projectile executes two different velocities for two different motion (horizontal motion and vertical motion).
• Acceleration: It is the change in velocity of the projectile in unit time. Though a particle executing two-dimensional motion has two different accelerations but usually, on Earth, a projectile has only one acceleration component. As the object is pulled downward by the gravitational force of the Earth, the vertical motion of the object experiences an acceleration towards the downward direction. However, as no force is given to the object externally, once it is thrown, it does not have any acceleration or force in its horizontal motion. Hence, there acts no acceleration for the horizontal motion of the projectile.

The displacement of a particle or object in one dimension (one direction) with respect to time is given by:

 

d = s₁t + \(\frac{1}{2}\)at²

 

where,

 

d = Displacement of the object

 

s₁ = Initial velocity of the object

 

a = Acceleration of the object

 

t = Time Period of motion

If any physical quantity ‘a’ varies with respect to some other physical quantity ‘b’, then it could be stated mathematically, that ‘a’ is a function of ‘b’. In mathematical notation, it could be written as:

 

a = f(b)

 

The derivative of the above equation would be the slope or gradient of the curve [a = f(b)] at any point (b). The zero of the derivative would be the values of (b) where the gradient of the curve would be zero. As at the maximum or minimum value of a function, the slope of the curve would be zero, hence, the value of the independent or varying variable (which is ‘b’ in this case) for which the gradient of the curve would be zero is considered as maxima or minima of the curve (function). To verify whether the value of ‘b’ is corresponding to maximum value of the curve or the minimum value of the curve, the equation of the curve [a = f(b)] is double differentiated. If the value of the double derivative at that particular value of independent variable ‘b’ is positive, then the obtained value of ‘b’ would be local minimum and if the value of the double derivative at that particular value of independent variable ‘b’ is negative, then the obtained value of ‘b’ would be local maximum.

It is the opposition or resistance offered by air towards any object executing motion. It is also known as drag. In case of linear air resistance, the force by which the ongoing motion is opposed by the air is given by:

 

F = - pv

 

Where,

 

F = Air resistance

 

p = Drag coefficient

 

v = velocity of motion

 

In case of quadratic air resistance, the force by which the ongoing motion is opposed by the air is given by:

 

F = - pv²

 

Where,

 

F = Air resistance

 

p = Drag coefficient

 

v = velocity of motion

As the shot put is thrown in an open field, it executes motion in both horizontal axis (or direction) as well as vertical axis (or direction). In this exploration, both the motion would be considered separately with suffix notations mentioned below:

Let, the initial velocity at which the shot put at which it has been thrown be s₁ making an angle of α with the horizontal axis. The horizontal component of velocity which initiates the horizontal motion is s_1h and the vertical component of velocity which initiates the vertical motion is s_1v

 

Therefore:

To represent the vertical and horizontal components of velocity in terms of the initial velocity of the throw and angle, certain trigonometric ratios have been used:

In figure 3:

 

cos cos α = \(\frac{OA}{OC}=\frac{\vec{s}_{1h}}{\vec s_1}\)

 

\(\vec s_{1h}=\vec s_{1}\times cos\ cos\ α\) ...(equation - 1)

 

sin sin α = \(\frac{AC}{OC}=\frac{\vec{s}_{1v}}{\vec s_1}\)

 

\(\vec s_{1v}=\vec s_{1}\times sin\ sin\ α\) ...(equation - 2)

 

From the displacement – time equation as mentioned in the background information section (Refer to Page 3), the displacement time equation for the motion of the shot put in the horizontal direction could be written as:

 

d_h = s_1h t + \(\frac{1}{2}\)a_h t²…(equation - 3)

 

In the field of shot put, as there is no acceleration component for the horizontal motion of the shot put,

 

a_h = 0

 

The equation (3) could be reframed as:

 

d_h = s_1h t + \(\frac{1}{2}\) (0)t²

 

d_h = s₁t × cos cos α (equation - 4)

 

The displacement time equation for the motion of the shot put in the vertical direction could be written as:

 

 d_v = s_1v t + \(\frac{1}{2}\)a_v t²…(equation - 5)

 

Due to the force of gravity, every object is pulled towards the center of the Earth with an acceleration of 9.81 m∙s^-2, the acceleration of the shot put in vertical direction could be written as:

 

a_v = - 9.81

 

The negative sign is due to the direction of the acceleration. As acceleration is a vector quantity, downward motion, is by convention assumed to be negatively directed. As the acceleration due to gravity is acting in downward direction, mathematically, the acceleration is written as negative.

 

The equation (5) could be reframed as:

 

d_v = s_1v t + \(\frac{1}{2}\) ( - 9.81)t²

 

d_v = s_1v t - \(\frac{1}{2}\) ( - 9.81) gt²

 

where,

 

g = acceleration due to gravity = - 9.81

 

dv = s₁t × sin sin α - \(\frac{1}{2}\) gt² …(equation - 6)

From equation (4):

 

d_h = s₁t × cos cos α

 

\(t=\frac{d_h}{s_1\ coscos\ α}\)

 

Plugging in the equation of time (t), in equation (6):

 

d_v = s₁ × \(\frac{x}{s_1\ coscos\ α}\)× sin sin α - \(\frac{1}{2}g\bigg(\frac{d_h}{s_1\ coscos\ α}\bigg)^2\)

 

As the equation of trajectory would be plotted in cartesian plane, let, the displacement in vertical axis be considered as y, and the displacement in the horizontal axis be considered as x:

 

 y = s₁ × \(\frac{x}{s_1\ coscos\ α}\)× sin sin α - \(\frac{1}{2}g\bigg(\frac{d_h}{s_1\ coscos\ α}\bigg)^2\)

 

y = x tan tan α - \(\frac{gx^2}{2s_1^2α}\)…(equation - 7)

 

In this exploration, as a comparative analysis would be developed between the trajectory of shot put in absence and presence of air resistance, thus, all the other parameters which could affect the trajectory should be controlled. There are two parameters which would affect the trajectory of motion apart from the wind current. They are – initial velocity of projection and angle of projection with respect to horizontal axis. As the velocity of throw depends on the athlete, so more the velocity, longer would be the distance traversed by the shotput. However, there would be only one angle at which horizontal displacement would be maximum.

The displacement time equation for horizontal motion was obtained as:

 

d_h = s₁t × cos cos α

 

The displacement time equation for vertical motion was obtained as:

 

d_v = s₁t × sin sin α  - \(\frac{1}{2}\)gt²

 

When the shot put will land on the ground after traversing the horizontal distance, the net displacement in the vertical direction would be zero.

 

Thus, the time period of the flight could be determined as:

 

0 = s₁t × sin sin α  - \(\frac{1}{2}\)gt²

 

\(\frac{1}{2}\)gt² - s₁t × sin sin α = 0

 

Multiplying ‘2’ in both the sides of the equation:

 

2 × \(\frac{1}{2}\)gt² - 2 × s₁t × sin sin α =2 × 0

 

gt² - 2 × s₁t × sin sin α = 0

 

Taking ‘t’ common from the left-hand side of the equation:

 

t[gt - 2s₁ × sin sin α] = 0

 

Either:

 

t = 0

 

Or:

 

gt - 2s₁ × sin sin α = 0

 

\(t=\frac{2s_1×sinsin\ α}{g}\)

 

As the position of the shot put before the throw (t = 0) was at the ground, the displacement of the shot put in vertical direction was zero. Hence, such a result has been obtained. However, the total time of flight will be:

 

\(t=\frac{2s_1×sinsin\ α}{g}\)...(equation - 8)

 

The distance traversed by the shot put in horizontal direction in this total time period would be:

 

d_h = s₁t × cos cos α

 

d_h = s₁ × \(\frac{2s_1×sinsin\ α}{g}\) × cos cos α

 

d_h = \(\frac{s_1^2sinsin\ 2α}{g}\) …(equation - 9)

 

Where,

 

2 sin sin α cos cos α = sin sin 2α

 

Applying the concepts of local maximum and minimum as discussed in the background information, the differentiation of equation (8) with respect to angle (α) would be:

 

\(\frac{d(d_h)}{dα}=\frac{d}{da}\bigg(\frac{s^2_1sinsin\ 2α}{g}\bigg)\)

 

As, s₁ is a constant value and does not depend on the angle, however, throughout the motion, the velocity changes, but the initial velocity cannot be changed as it is the velocity with which the athlete throws the shot put and g is also a constant value as it is the acceleration due to gravity, there two parameters would be taken out of the derivative:

 

\(\frac{d(d_h)}{dα}=\frac{s^2_1}{g}× \frac{d}{dα}\)(sin sin 2α)

 

 \(\frac{d(d_h)}{dα}=\frac{s^2_1}{g}×\bigg[\frac{d}{2(dα)}(sin\ sin\ 2α)×\frac{d}{(dα)}(2α)\bigg]\)

 

\(\frac{d(d_h)}{dα}=\frac{s^2_1}{g}\) × [cos cos 2α ×2]

 

\(\frac{d(d_h)}{dα}=\frac{2s^2_1coscos\ 2α}{g}\) …(equation - 10)

 

Equating the derivative to zero:

 

\(\frac{2s^2_1cos\ cos\ 2α}{g}=0\)

 

cos cos 2α = 0

 

 2α = (2n - 1)\(\frac{\pi}{2}\)

 

Since, α is an acute angle, n = 1

 

2α = \(\frac{\pi}{2}\)

 

α = \(\frac{\pi}{4}\)

 

To verify the value of α being corresponding to maximum or minimum value of displacement, equation (9) is again differentiated with respect to α:

 

\(\frac{d^2(d_h)}{dα^2}=\frac{d}{dα}\bigg(\frac{2s^2_1coscos\ 2α}{g}\bigg)\)

 

\(\frac{d^2(d_h)}{dα^2}=\frac{2s^2_1}{g}×\frac{d}{dα}\) (cos cos 2α)

 

\(\frac{d^2(d_h)}{dα^2}=\frac{2s^2_1}{g}×\bigg[\frac{d}{d(2α)}(cos\ cos\ α2)×\frac{d}{d(α)}(2α)\bigg]\)

 

\(\frac{d^2(d_h)}{dα^2}=\frac{2s^2_1}{g}×[-sin\ sin\ 2α×2]\)

 

\(\frac{d^2(d_h)}{dα^2}=-\frac{4s^2_1sin\ sin\ 2α}{g}\)

 

\(\frac{d^2(d_h)}{dα^2}|_{α=\frac{\pi}{4}}=-\frac{4s^2_1sin\ sin\ 2×\frac{\pi}{4}}{g}\)

 

\(\frac{d^2(d_h)}{dα^2}|_{α=\frac{\pi}{4}}=-\frac{4s^2_1×1}{g}\)

 

\(\frac{d^2(d_h)}{dα^2}|_{α=\frac{\pi}{4}}=-\frac{4s^2_1}{g}\)

 

Since, the value of the double derivative is negative at \(α=\frac{\pi}{4},\) the displacement would be maximum, at \(α=\frac{\pi}{4}.\)

From equation (7):

 

y = x tan tan α - \(\frac{gx^2}{2s^2_1α}\)

 

Plugging in the value of \(α=\frac{\pi}{4}\) in equation (7):

 

y = x tan tan \(\bigg(\frac{\pi}{4}\bigg)-\frac{gx^2}{2s^2_1\frac{1}{2}}\)

 

y = x × 1 - \(\frac{gx^2}{2s^2_1×\frac{1}{2}}\)

 

y = x  - \(\frac{gx^2}{s^2_1}\) ...(equation - 11)

As the air resistance considered in this exploration is quadratic in nature, the expression of the air resistance could be written as:

 

F_air = - ks₁²

 

Where,

 

k = Coefficient of the air resistance

 

s₁= Initial velocity of shot put

 

As the air resistance is opposing the motion, thus, negative sign has been introduced in the above equation. Let, the wind is blowing making an angle of β with the horizontal axis:

 

Therefore, as velocity vectors have been resolved in page 5 and 6, similarly, the air resistance (force) vector would be resolved into:

 

F_air|h = - ks₂² cos cos β

 

F_air|h = - ks_2h²

 

Where,

 

s_2h² = s₂² cos cos β

 

F_air|v = - ks₂² sin sin β

 

F_air|v = - ks_2v²

 

Where,

 

s_2v² = s₂² sin sin β

 

F_air|h = Horizontal component of air resistance

 

F_air|v = Vertical component of air resistance

 

s₂ = Final velocity of shot put with air resistance

 

s_2h = Horizontal velocity component of shot put with air resistance

 

s_2v = Vertical velocity component of shot put with air resistance

 

From the Newton’s second law of motion, it could be said that, the net force acting on any object is equal to the product of mass of the object and the acceleration of the object.

F_net|h = mα_h

 

Where,

 

F_net|h = Total force acting in horizontal motion

 

m = mass of the shot put

 

a_h = acceleration of shot put in horizontal direction after air resistance

 

F_net|h = ma_h

 

- ks_2h² = ma_h

 

- ks_2h² = m\(\frac{d(s_{2h})}{dt}\)

 

\(-\frac{k(dt)}{m}=\frac{d(s_{2h})}{s_{2h}\ ^2}\)

 

Integrating the above equation:

 

\(\displaystyle\int-\frac{k(dt)}{m}=\displaystyle\int\frac{d(s_{2h})}{s_{2h}\ ^2}\)

 

As the drag coefficient (k), mass of the shot put (m) and the angle of wind (β) are constant parameters, those will come out of the integration:

 

\(-\frac{k}{m}×t=\frac{s_{2h}^\ \ ^{-2+1}}{-2+1}+c\)

 

\(-\frac{k}{m}×t=-\frac{1}{s_{2h}}+c\)

 

\(\frac{kt}{m}=\frac{1}{s_{2h}}+c\)

 

Where,

 

c = integration constant

 

At the initial condition, (t = 0),

 

s_2h = s_1h = s₁ cos cos α

 

This is because, when the shot put has not been thrown or lifted above the ground, there would not be any effect of air resistance.

 

\(\frac{k(0)}{m}=\frac{1}{s_1\ coscos\ α}+c\)

 

\(c=-\frac{1}{s_1\ coscos\ α}\)

 

Therefore, plugging in the value of (c) into the main equation:

 

\(\frac{kt}{m}=\frac{1}{s_{2h}}-\frac{1}{s_1\ coscos\ α}\)

 

\(\frac{1}{s_{2h}}=\frac{1}{s_1\ coscos\ α}+\frac{kt}{m}\)

 

\(\frac{1}{s_{2h}}=\frac{m+kts_1\ coscos\ α}{s_1\ mcoscos\ α}\)

 

\(s_{2h}=\frac{s_1\ mcoscos\ α}{m+kts_1\ coscos\ α}\) …(equation - 12)

 

Displacement of the shot put in horizontal direction in presence of air resistance:

 

\(d_h=\displaystyle\int s_{2h}dt\)

 

\(d_h\displaystyle\int\frac{s_1\ mcoscos\ α}{m+kts_1\ coscos\ α}dt\)

 

By integrating the above integral by the method of substitution, Let’s assume that:

 

u = m + kts₁ cos cos α

 

Differentiating the above function (u) with respect to time (t):

 

\(\frac{du}{dt}\) = 0 + ks₁ cos cos α

 

dt = \(\frac{du}{ks_1\ coscos\ α }\)

 

Plugging in the value of (dt) and (u) in the above integral:

 

\(d_h=\displaystyle\int\frac{s_1mcoscos\ α}{u}\times\frac{du}{ks_1coscos\ α }\)

 

\(d_h=\displaystyle\int\frac{m}{u}×\frac{du}{k}\)

 

\(d_h=\displaystyle\int\frac{m}{k}×\displaystyle\int\frac{1}{u}du\)

 

d_h = \(\frac{m}{k}\) × ln ln u + c₁

 

Where,

 

c₁ = integral constant

 

d_h = \(\frac{m}{k}\)× ln ln (m + kts₁ cos cos α) + c₁ ...(equation - 13)

 

At the initial condition (t = 0), the displacement in horizontal axis would also be zero,

 

Therefore:

 

0 = \(\frac{m}{k}\)× ln ln (m + k (0) s₁ cos cos α) + c₁ = - \(\frac{m}{k}\)ln ln (m)

 

Plugging in the value of c₁ in equation (13):

 

d_h = \(\frac{m}{k}\) × ln ln (m + kts₁ cos cos α) - \(\frac{m}{k}\) ln ln (m)

 

d_h = \(\frac{m}{k}\)[ ln ln (m + kts₁ cos cos α) - ln ln (m) ]...(equation - 14)

F_net|v - mg = ma_v

 

Where,

 

F_net|v = Total force acting in vertical motion

 

m = mass of the shot put

 

a_v = acceleration of shot put in vertical direction after air resistance

 

F_net|v - mg = mav

 

- ks_2v² - mg = ma_v

 

- ks_2v² - mg = m\(\frac{d(s_{2v})}{dt}\)

 

\(\frac{dt}{m}=\frac{1}{-ks_{2v}\ ^2-mg}d(s_{2v})\)

 

\(-\frac{dt}{m}=\frac{1}{-ks_{2v}\ ^2+mg}d(s_{2v})\)

 

Integrating both sides of the equation with respect to s_2v:

 

\(\displaystyle\int-\frac{dt}{m}=\displaystyle\int\frac{1}{ks_{2v}\ ^2+mg}d(s_{2v})\)

 

Let, the integral in the right-hand side be solved separately:

 

\(\displaystyle\int\frac{1}{ks_{2v}\ ^2+mg}d(s_{2v})\)

 

\(=\displaystyle\int\frac{1}{k\bigg(s_{2v}\ ^2+\frac{mg}{k}\bigg)}d(s_{2v})\)

 

\(=\frac{1}{2}\displaystyle\int\frac{1}{\bigg(s_{2v}\ ^2+\frac{mg}{k}\bigg)}d(s_{2v})\)

 

Let, \(\frac{mg}{k}\) = α²

 

\(=\frac{1}{k}\displaystyle\int\frac{1}{\bigg(s_{2v}\ ^2+α^2\bigg)}d(s_{2v})\)

 

\(=\frac{1}{k}\displaystyle\int\frac{\frac{1}{α^2}}{\bigg(\frac{s_{2v}\ ^2}{α^2}+\frac{α^2}{α^2}\bigg)}d(s_{2v})\)

 

\(=\frac{1}{α^2k}\displaystyle\int\frac{1}{\bigg((\frac{s_{2v}}{α})^2+1\bigg)}d(s_{2v})\)

 

By the method of substitution of integration:

 

\(u=\frac{s_{2v}}{α}\)

 

Differentiating both sides of the above equation with respect to s_2v:

 

\(\frac{du}{ds_{2v}}=\frac{1}{α}\)

 

ds_2v = α du

 

Plugging in the values of the above equations in the principle integral:

 

   \(\frac{1}{α^2k}\displaystyle\int\frac{1}{\bigg((\frac{s_{2v}}{α})^2+1\bigg)}d(s_{2v})\)

 

\(=\frac{1}{α^2k}\displaystyle\int\frac{1}{(u^2+1)}α\ du\)

 

\(=\frac{1}{α^2k}\displaystyle\int\frac{1}{(u^2+1)} du\)

 

    = \(\frac{1}{ak}\) × u + c₂

 

Where,

 

c₂ = integral constant

 

\(\frac{1}{ak}×\bigg(\frac{s_{2v}}{α}\bigg)+c_2\)

 

\(=\frac{1}{\sqrt{\frac{mg}{k}k}}×\bigg(\frac{s_{2v}}{\sqrt{\frac{mg}{k}}}\bigg)+c_2\)

 

\(=\frac{1}{\sqrt{{mgk}}}×\bigg(s_{2v\sqrt{\frac{k}{mg}}}\bigg)+c_2\)

 

Therefore:

 

\(\displaystyle\int-\frac{dt}{m}=\displaystyle\int\frac{1}{ks_{2v}\ ^2+mg}d(s_{2v})\)

 

\(-\frac{t}{m}=\frac{1}{\sqrt{mgk}}×\bigg(s_{2v}\sqrt{\frac{k}{mg}}\bigg)\) + c₂ …(equation - 15)

 

At initial conditions, (t = 0):

 

s_2v = s_1v = s₁ sin sin α

 

This is because, when the shot put has not been thrown or lifted above the ground, there would not be any effect of air resistance.

 

\(-\frac{0}{m}=\frac{1}{\sqrt{mgk}}×\bigg(s_1\ sin\ sin\ α\sqrt{\frac{k}{mg}}\bigg)\) + c₂

 

\(c_2=-\frac{1}{\sqrt{mgk}}×\bigg(s_1\ sin\ sin\ α\sqrt{\frac{k}{mg}}\bigg)\)

 

Therefore, the plugging in the value of c₂ in equation (15):

 

\(-\frac{t}{m}=\frac{1}{\sqrt{mgk}}×\bigg(s_{2v}\sqrt{\frac{k}{mg}}\bigg)-\frac{1}{\sqrt{\frac{k}{mg}}}×\bigg(s_1\ sin\ sin\ α\sqrt{\frac{k}{mg}}\bigg)\)

 

\(\frac{1}{\sqrt{mgk}}×\bigg(s_{2v}\sqrt{\frac{k}{mg}}\bigg)=\frac{1}{\sqrt{mgk}}×\bigg(s_1\ sin\ sin \ α\sqrt{\frac{k}{mg}}\bigg)-\frac{t}{m}\)

 

\(\bigg(s_{2v}\sqrt{\frac{k}{mg}}\bigg)=\bigg(s_1\ sin\ sin\ α\sqrt{\frac{k}{mg}}\bigg)-\frac{t\sqrt{mgk}}{m}\)

 

\(s_{2v}=\sqrt{\frac{k}{mg}}=tan\ tan\bigg[\bigg(s_1\ sin\ sin \ α\sqrt{\frac{k}{mg}}\bigg)-\frac{t\sqrt{mgk}}{m}\bigg]\)

 

\(s_{2v}=\sqrt{\frac{mg}{k}}tan\ tan\bigg[\bigg(s_1\ sin\ sin \ α\sqrt{\frac{k}{mg}}\bigg)-\frac{t\sqrt{mgk}}{m}\bigg]\) ... (equation - 16)

 

Displacement of the shot put in horizontal direction in presence of air resistance:

 

\(d_v =\displaystyle\int s_{2v}dt\)

 

\(d_v=\displaystyle\int \bigg[\sqrt{\frac{mg}{k}}tan\ tan\bigg[\bigg(s_1\ sin\ sin\ α\sqrt{\frac{k}{mg}}\bigg)-\frac{t\sqrt{mgk}}{m}\bigg]\bigg]dt\)

 

\(d_v=\sqrt{\frac{mg}{k}}\displaystyle\int \bigg[tan\ tan\bigg[\bigg(s_1\ sin\ sin\ α\sqrt{\frac{k}{mg}}\bigg)-\frac{t\sqrt{mgk}}{m}\bigg]\bigg]dt\)

 

\(d_v=\sqrt{\frac{mg}{k}}\displaystyle\int \frac{tantan\bigg\{\bigg(s_1\ sinsin\ α\sqrt{\frac{k}{mg}}\bigg)\bigg\}-tantan\bigg\{\frac{t\sqrt{mgk}}{m}\bigg\}}{1+tantan\bigg\{\bigg(s_1\ sinsin\ α\sqrt{\frac{k}{mg}}\bigg)\bigg\}-tantan\bigg\{\frac{t\sqrt{mgk}}{m}\bigg\}}\)

 

\(d_v=\sqrt{\frac{mg}{k}}\displaystyle\int \frac{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}-tantan\bigg\{\frac{t\sqrt{mgk}}{m}\bigg\}}{1+\bigg\{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}\bigg\}tantan\bigg\{\frac{t\sqrt{mgk}}{m}\bigg\}}\) …(equation - 17)

 

To solve the integral at the right-hand side of the equation (17), let’s solve the integral separately:

 

\(\displaystyle\int \frac{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}-tantan\bigg\{\frac{t\sqrt{mgk}}{m}\bigg\}}{1+\bigg\{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}\bigg\}tantan\bigg\{\frac{t\sqrt{mgk}}{m}\bigg\}}\)

 

Let, the following terms be assumed as different variables for simplication:

 

p = s₁ sin sin α \(\sqrt{\frac{k}{mg}}\)

 

q = \(\frac{\sqrt{mgk}}{m}\)

 

\(\displaystyle\int\frac{p-tantan(qt)}{1+p×tantan(qt)}dt\)

 

\(\frac{\{p\ coscos\ (qt)-sinsin\ (qt)\}t}{p\ sinsin\ (qt)+coscos\ (qt)}\) + c₅

 

where,

 

c₅ = integration constant

 

\(=\frac{\big\{s_1sinsin\ α\sqrt{\frac{k}{mg}}coscos\big(\frac{\sqrt{mgk}}{m}t\big)-sinsin\big(\frac{\sqrt{mgk}}{m}t\big)\big\}}{s_1sinsin\ α\sqrt{\frac{k}{mg}}sinsin\big(\frac{\sqrt{mgk}}{m}t\big)+coscos\big(\frac{\sqrt{mgk}}{m}t\big)}\) + c₅

From equation (17):

 

\(d_v=\sqrt{\frac{mg}{k}}\displaystyle\int \frac{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}-tantan\bigg\{\frac{t\sqrt{mgk}}{m}\bigg\}}{1+\bigg\{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}\bigg\}tantan\bigg\{\frac{t\sqrt{mgk}}{m}\bigg\}}\)

 

\(d_v=\sqrt{\frac{mg}{k}}\bigg[\frac{\big\{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}coscos\bigg(\frac{\sqrt{mgk}}{m}t\bigg)-sinsin\bigg(\frac{\sqrt{mgk}}{m}t\bigg)\bigg\}t}{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}sinsin\bigg(\frac{\sqrt{mgk}}{m}t\bigg)+coscos\bigg(\frac{\sqrt{mgk}}{m}t\bigg)}+c_5\bigg]\) …(equation - 18)

 

At initial conditions (t = 0),

 

d_v = 0

 

Therefore:

 

\(0=\sqrt{\frac{mg}{k}}\bigg[\frac{\big\{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}coscos\bigg(\frac{\sqrt{mgk}}{m}(0)\bigg)-sinsin\bigg(\frac{\sqrt{mgk}}{m}(0)\bigg)\bigg\}(0)}{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}sinsin\bigg(\frac{\sqrt{mgk}}{m}(0)+coscos\bigg(\frac{\sqrt{mgk}}{m}(0)\bigg)}+c_5\bigg]\)

 

c₅ = 0

 

Plugging in the value of c₅, in equation (18):

 

\(d_v=\sqrt{\frac{mg}{k}}\bigg[\frac{\big\{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}coscos\bigg(\frac{\sqrt{mgk}}{m}t\bigg)-sinsin\bigg(\frac{\sqrt{mgk}}{m}t\bigg)\bigg\}t}{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}sinsin\bigg(\frac{\sqrt{mgk}}{m}t\bigg)+coscos\bigg(\frac{\sqrt{mgk}}{m}t\bigg)}\bigg]\)…(equation - 19)

 

SUMMARY OF THE ABOVE SECTION:

 

Motion of shot put in absence of air resistance:

 

Displacement in vertical direction (d_v) = s₁t × sin sin α - \(\frac{1}{2}\)gt²

 

Displacement in horizontal direction (d_v) = s₁t × cos cos α

 

Motion of shot put in presence of air resistance:

 

Displacement in vertical direction (d_v)

 

\(=\sqrt{\frac{mg}{k}}\bigg[\frac{\big\{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}coscos\bigg(\frac{\sqrt{mgk}}{m}t\bigg)-sinsin\bigg(\frac{\sqrt{mgk}}{m}t\bigg)\bigg\}t}{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}sinsin\bigg(\frac{\sqrt{mgk}}{m}t\bigg)+coscos\bigg(\frac{\sqrt{mgk}}{m}t\bigg)}\bigg]\)

 

Displacement in horizontal direction (d_v)

 

\(=\frac{m}{k}[ln\ ln(m+kts_1\ cos\ cos\ α)-ln\ ln(m)]\)

In this section of the internal assessment, the effect of air resistance has been graphically presented by taking a case with particular values of the parameters in the equations of displacement both in horizontal as well as vertical direction.

 

From the above exploration, it has been studied that, an angle of 45° should be the angle of projection for having maximum displacement in horizontal axis. As, achieving a maximum horizontal displacement is the ultimate goal of the game of shot put, the angle of projection in this exploration would be considered as 45°. In this exploration, only the case study for the game play of men has been considered as otherwise, another case study for the game play of women would be repetitive. Hence, the mass of the shot put has been considered as 7.26 kg. The average velocity of throw of shot put for men has been recorded as 10.24 m∙s^-1. Let us consider, that the coefficient of drag (k) be 0.5.

 

From the above data:

 

Mass of shot put(m) = 7.26 kg

 

Initial velocity of throw (s₁) = 10.24 m∙s^-1

 

Coefficient of drag = 0.5

 

Angle of projection (α) = 45°

d_v = s₁t × sin sin α - \(\frac{1}{2}\) gt²

 

= (10.24)(0) × sin sin (45°) - \(\frac{1}{2}\) (9.81)(0)² = 0

 

d_h = s₁t × cos cos α

 

=10.24(0) × cos cos (45°) = 0

d_v = \(\sqrt{\frac{7.26×9.81}{0.5}}×\)

 

\(\bigg[\frac{\bigg\{s_1sinsin\ (45°)\sqrt{\frac{0.5}{7.26×9.81}}coscos\big(\frac{\sqrt{7.26×9.81×0.5}}{7.26}(0)\big)-sinsin\big(\frac{\sqrt{7.26×9.81×0.5}}{7.26}(0)\big)\bigg\}(0)}{s_1sinsin\ (45°)\sqrt{\frac{0.5}{7.26×9.81}}sinsin\big(\frac{\sqrt{7.26×9.81×0.5}}{7.26}(0)\big)+coscos\big(\frac{\sqrt{7.26×9.81×0.5}}{7.26}(0)\big)\bigg\}}\bigg]\)

 

= 0

 

d_h = \(\frac{7.26}{0.5}\) [ln ln (7.26 + 0.5(0)(10.24) cos cos (45°) ) - ln ln (7.26) ] = 0

In the above graph, the horizontal displacement of the shot put in the horizontal direction has been plotted along the X – Axis and the vertical displacement has been plotted along the Y – Axis. It is clearly observed that if there is an air resistance (quadratic in nature) with a drag coefficient of 0.5 kg∙m^-1∙s, there would be a significant change in the horizontal distance covered by the shot put. If the shot put has been thrown at an initial velocity of 10.24 m∙s^-1, at an angle of elevation with respect to the horizontal axis of 45°, there would be a difference in horizontal distance covered of 6.52 m, since the displacement of the shot put in absence of air resistance was around 10.70 m and that of in presence of air resistance was approximately 4.18 m.

 

Hence, air resistance may cause significant error in judgement in any game where shot put is an event of occurrence. Thus, whenever the game of shotput would be played, it should be made sure that there is no breeze blowing in the field otherwise a player would be benefitted over other depending on the presence of wind.

Investigating the effect of quadratic air resistance (wind current) on the trajectory of a shot put (projectile) and compare that with respect to a controlled state (absence of air resistance).

 

There is a significant impact in the trajectory and hence the result of shot put (both male and female) in presence of quadratic air resistance compared with the trajectory in absence of air resistance.

• The obtained equation to account the horizontal displacement of shot put in absence of air resistance is: s₁t × cos cos α, where, s₁ is the initial velocity of throw (in m.s^-1), t is the time (in sec), α is the angle of projection with respect to horizontal axis.
• The obtained equation to account the vertical displacement of shot put in absence of air resistance is: s₁t × sin sin α - \(\frac{1}{2}\) gt², where, s₁ is the initial velocity of throw (in m.s^-1), t is the time (in sec), α is the angle of projection with respect to horizontal axis, and g is the acceleration due to gravity (in m.s^-2).
• From the above two displacement equations, it could be stated that the mass of shot put does not affect the displacement or trajectory of motion in absence of air. Hence, for both the variants of mass of shot put (for male and female), there would not be any change in trajectory if the initial velocity of throw and the angle of projection are the same.
• The obtained equation to account the horizontal displacement of shot put in absence of air resistance is: \(\frac{m}{k}[ln\ ln(m+kts_1\ cos\ cos\ α)-ln\ ln(m)]\) where, s₁ is the initial velocity of throw (in m.s^-1), t is the time (in sec), α is the angle of projection with respect to horizontal axis, m is the mass of the shot put (in kg) and k is the drag coefficient.
• The obtained equation to account the vertical displacement of shot put in absence of air resistance is: \(\sqrt{\frac{mg}{k}}\bigg[\frac{\big\{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}coscos\bigg(\frac{\sqrt{mgk}}{m}t\bigg)-sinsin\bigg(\frac{\sqrt{mgk}}{m}t\bigg)\bigg\}t}{s_1\ sinsin\ α\sqrt{\frac{k}{mg}}sinsin\bigg(\frac{\sqrt{mgk}}{m}t\bigg)+coscos\bigg(\frac{\sqrt{mgk}}{m}t\bigg)}\bigg],\) where, s₁ is the initial velocity of throw (in m.s^-1), t is the time (in sec), α is the angle of projection with respect to horizontal axis, g is the acceleration due to gravity (in m.s^-2),  m is the mass of the shot put (in kg) and k is the drag coefficient.
• For the motion in presence of air resistance, it could be stated in reference with the two displacement equations of motion that with increase in mass of the shot put, the both the horizontal and vertical displacement would decrease. Similarly, with an increase in drag coefficient, both the horizontal and vertical displacements would decrease.
• For maximum displacement of projectile, the object should be thrown at an angle of 45° with respect to the horizontal axis.

• The data collected for the case study as well as for research purpose are taken either from educational websites or sporting event websites. Hence, the data used in the case study could be considered as authentic which made the exploration more coherent and error-free.
• The variation in trajectory in presence and absence of air resistance was not only described by deriving equations, but also by graphical representation. It has made the exploration easier and more convenient for presentation and also to give a generic idea to the sport organizing committee and take necessary steps.
• A case study with real data has been developed.

• A height at which the shot put is released during the throw has considered to be zero in the case study which is not true. As described in the background information, the shot put is thrown from over the shoulder. However, as there was no data available in the open sources regarding the height from which the shot put is thrown, it was considered to be zero.
• The case study that has been undertaken has for the game play of men as the mass of shot put considered was 7.26 kg. However, it was already been studied that change in mass affects the trajectory of the shotput. However, another case study for shot put played by women was not done to avoid repetition of mathematical processing.
• It has been studied that usually, the shot put was thrown at an angle of 37° with respect to the horizontal axis. However, in the case study, it was considered as 45°. This is because, the case study was done with data offering an optimum range or displacement in the horizontal axis.

In this exploration, the effect of quadratic air resistance on the trajectory of shot put was investigated. However, air resistance could be present in the field in linear form as well. In that scenario, the trajectory of the shot put would be different from what has been observed in this exploration. Thus, I personally believe and would like to extend this exploration and investigate the effect of linear air resistance on trajectory of a shot put (both male and female) and compare that with respect to the trajectory of no air resistance condition (control state). The methodology of the exploration would be similar to this investigation; however, while interpreting the effect of air resistance (drag force), the general equation should be considered linear:

 

F = - kv

 

Where, k would be the coefficient of drag and v would be the velocity of the shot put.

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