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My favorite subject is science. I find it very interesting. Science can be applied everywhere in our day to day life and this is what makes it interesting for me.

I am a sports jock and have always taken interest in it. I have participated in school sports even from the very 1st grade. I play football quite well. I am in the school's football team as well. But it is not only football that I play, I tend to play any sport that I get a chance.

For this year's sports, a friend of mine gave his name for javelin throw. I accompanied him to his practice where I tried playing it as well. I quite enjoyed it and somehow my friend convinced me to give my name for javelin throw in the sport's day.

I take every sport seriously. I try my best to play well and so I could not take this casually. I started practicing for javelin throw. A game is not just learning to play on the field, there is a lot theory behind it. Understanding and applying it, is very important to perfect a game.

I had to do the same with javelin throw. Field practice would be perfect only if I could master the theoretical concept. So, I started finding out information about this sport. There was science to it and I read all available information with great interest. I came to know about the spear and quite a number of tips about the game, but I could not get the answer to the question that bothered me the most.

While practicing on field, I at times found it difficult to manage during wind and I also had no idea about the angle of projection. The question that I was trying to find was that to what extent the angle of projection of spear during javelin throw and wind speed acting in horizontal direction as that of the projection of the spear, affect the range of the projectile.

So, I have decided to research on the above stated question and this is what the IA is all about.

The main motive of this IA is to study the trajectory of Javelin throw and explore the effect of angle of projection of the throw in winning the game. As javelin is played in open field, air resistance and air current are two uncontrolled variable that may affect the game. Thus, another motive of this IA, in addition to the study of angle of projection, is to investigate the effect of wind current in a direction same as that of the direction of horizontal motion of javelin throw and that of the opposite direction as well. Finally, I will find the effect of air resistance in the trajectory of Javelin Throw.

To what extent does the angle of projection of javelin throw, wind speed acting in horizontal direction as that of the projection of javelin, and air resistance affect the range of the projectile (javelin)?

Javelin throw is one of the most watched and interesting games in athletics. The winning factor of the game is the distance the javelin has covered in a single throw. The one whose range of javelin throw is maximum, wins the game. In the game, there are several parameters on which the range of the javelin depends. It depends on the initial velocity of throw and also the direction of projection. We know that, maximum the initial velocity of throw will be, more the distance the javelin will cover but the effect of angle of projection will be studied in this IA. As, with stepper angle of projection, the javelin cannot cover a larger distance and same for flatter angle of projection. A few equations of motion in physics are very essential in this IA. The relationship between displacement and time will be extensively used in the process of calculation and thus it is shown below:

\(s = ut +\frac{1}{2}at^2\)………(1)

where, s is the displacement done by the body, u is the initial velocity of the body, a is acceleration of the body, and t is time.

As we know that, with increased velocity of javelin, the range of the projectile will also increase. As javelin throw game has been played in open ground, the air current may increase the velocity of the javelin which will earn more range and become a favorable factor for the player. Similarly, variable velocity may offer different aspects compared to the other players in the game. Furthermore, if the wind is blowing in opposite direction as that of the horizontal direction of javelin, then the velocity of javelin will decrease which will eventually decrease the range of javelin and being a disadvantage for the player. As a result, the evaluation of the game will be biased. This is due to relative velocity. So, in this IA, we will study the effect of wind current in velocity of javelin resulting in achieving different range. The basic formulation of relative velocity which will be essential in the proceedings of this IA are as follows:

If the direction of wind is exactly equal to that of the horizontal direction of javelin throw, then, v_{j} being the initial velocity of javelin and v_{w} be the velocity of wind, the effective velocity of javelin will be:

v_{eff} = v_{j} + v_{w}………(2)

Similarly, if the direction of wind is exactly opposite to that of the horizontal direction of javelin throw, then, v_{j} being the initial velocity of javelin and v_{w} be the velocity of wind, the effective velocity of javelin will be:

v_{eff} = |v_{j} - v_{w}|………(3)

Air Resistance is the resistive force that is experienced by any moving object through air medium. It is due to the friction that takes place between the air molecules and the moving object. The formula of Air Resistance is shown as follows:

\(F = \frac{\rho C_DA}{2}×v^2\)

where, F is the resistive force (Air Resistance), ρ is density of air, C_{D} is drag coefficient, A is the area of cross section of moving object, and v is velocity of the object.

**Derivation of Displacement – Time Equation**

In this IA, we will be using the equation (1) for derivation of the equation of range of trajectory of javelin. The derivation of equation (1) is shown below:

From the definition of velocity, we know that, “rate of change of displacement or the change of displacement of any body in unit time is velocity”. Moreover, acceleration is defined as, “rate of change of velocity”. Therefore, from the definition of acceleration, we can write,

\(a=\frac{v-u}{t}\)

where, a is the acceleration, u is the initial velocity, v is the final velocity and t is the time for which the body was in motion.

∴ v = u + at

Again, from the definition of velocity, we can write

\(v=\frac{S_f-s_i}{t}\)

where, s_{f} is the final displacement, s_{i} is the initial displacement of the body.

For very small instant of time, we can write,

\(\frac{ds}{dt}=v\)

\(=>\frac{ds}{dt}=u+at\)

=> ds = udt + atdt

\(=>\displaystyle\int ds = \displaystyle\int udt +\displaystyle\int atdt\)

\(=> s = ut + \frac{1}{2}\,at^2 +c\)

Here, c is the integration constant which can be defined as the initial displacement of the body. If there is no initial displacement, i.e., c = 0, then we can conclude:

\(s=ut+\frac{1}{2}at^2\)

In vertical axis, let the modified form of equation (1) be written as:

\(s_y=u_yt+\frac{1}{2}a_yt^2\)………(4)

In horizontal axis, let the modified form of equation (1) be written as:

\(s_x = u_x t + \frac{1}{2}a_xt^2\)………(5)

**Parametric values:** As the net displacement in vertical direction in a javelin throw is zero, and the acceleration is due to gravity, the respective values will be:

s_{y }= 0 m; a_{y }= - 9.81 m/s^{2}

Similarly, the net displacement in horizontal direction in a javelin throw is the range which is the dependent variable of this IA. Let the range be R, and the acceleration in horizontal direction is zero. Thus, the respective values will be:

S_{x} = R m; a_{x} = 0 m/s^{2}

**Resolution of vector:** Resolution of vector is a mathematical tool using which the horizontal component and the vertical component of any vector is found out. In case of javelin throw, it is thrown at an angle with some initial velocity; which implies that the initial velocity has a definite direction. If the initial velocity of projection makes an angle of \(\theta\) with respect to the horizontal axis, then:

From figure 1, blue arrow is the direction of projection thus, we can redraw the section to mathematically derive the horizontal and vertical component of velocity.

From figure 2, we can write,

\(sin \, \,sin\,\theta=\frac{AB}{OA}=\frac{OC}{OA}=\frac{u_y}{u}\)

=> uy = u× sin sin *θ** *………(6)

\(cos \, \, cos\,\theta=\frac{OB}{OA}=\frac{U_X}{U}\)

=> ux = u× cos cos ………(7)

**Derivation using concepts of solving simultaneous equation -**

Using equation (4), (6) and the parametric values, we can write,

\(sy = uy \, \,t + \frac{1}{2}ay \, \,gt^2\)

\(=> 0 = u× sin \, \,sin \, θ ×t - \frac{1}{2}\, gt^2\)

\(=> \biggl(2u × sin \, \,sin \, \theta×t - gt^2\biggl)= 0\)

=> (2*u × sin sin **θ **- gt*) × *t *= 0

∴ *t* = 0

∴ 2*u × sin sin** θ **- gt* = 0

\(=> t = \frac{2u\,×\,sin\,sin \,\theta }{g}\)

This implies, at these two-time instances, the vertical displacement will be zero. Thus, the total time of flight will be:

\(T = \frac{2u\,×\,sin\,sin \,\theta }{g}\)

Now using equation (5), (7) and the parametric values, we can write:

\(s_y = u_y t +\frac{1}{2}ayt^2\)

\(=> R = u_x T + \frac{1}{2}× (0) × T2× (0) × T^2\)

Here, R is the range and T is the total time period or total time of flight.

∴ R = u_{x }T

\(=> R = u× cos \, \, cos\, \, \theta × \frac{2u×sin sin \,\theta }{g}\)

\(=> R = \frac{u^2×2sin sin \,\theta\, coscos\, \theta }{g}\)

\(=> R = \frac{u^2×sin sin \,2\theta }{g}\)………(8)

**Maxima and Minima of calculus**

If a function f(x) is differentiated with respect to x, then the derivative will be the slope of the curve f(x). Now, if the derivative is equated to zero, it implies that the slope of the curve is a line which is parallel to the X – Axis because slope of a straight line can be defined as the tangent of angle that the straight line makes with positive X – Axis. Now, at two instances, the slope of a curve could be parallel to the horizontal axis. Those are, either the slope is at maximum value of the curve or at the minimum value of the curve. To determine that, the derivate is differentiated again with respect to x. If the value of the derivative is positive, then the obtained slope was at minimum value of the curve and if the value of the derivative is negative, then the obtained slope was at the maximum value of the curve.

**Investigation on the effect of angle of projection**

The match which I have seen, the velocity of projection of the athlete was 30 m/s.

Therefore, we can compute the value and establish an equation of range as follows:

\(R = \frac{u^2×sinsin\, 2\theta}{g}\)

\(R = \frac{30^2×sinsin\, 2\theta}{9.81}\)

\(=> R = \frac{900×sinsin\, 2\theta}{9.81}\)

R = 91.74 × sin 2 ………(9)

Now, we will use Desmos to show graphically the variation of range due to angle or projection.

From the graph, we can find that the range of javelin depends on the angle of projection in sinusoidal nature, provided the angle should be acute.

To find the maximum range, the concept of maxima and minima will be used. I will differentiate the equation no (9) with respect to \(\theta\), to find the value of angle of projection for which the range will be maximum.

*R* = 91.74× *sin sin* 2θ

\(\frac{dR}{d\theta}=\frac{d}{d\theta}(91.74× sin \, \,sin \, \, 2θ )\)

\(=> \frac{dR}{d\theta} = 91.74×\frac{d}{d\theta}(sin \, \,sin \, \,2θ )\)

\(=> \frac{dR}{d\theta}= 91.74× 2× cos \, \,cos \, 2θ\)

=> 0 = 183.48× *cos cos* 2θ

=>* cos cos* 2θ = 0

\(=> 2θ = \frac{\pi}{2}\)

\(=> θ = \frac{\pi}{4}\)

Therefore, for maximum range, the angle of projection should be \(\frac{\pi}{4}\) or 45°

**Investigation on the effect of wind speed**

To establish the effect of wind in Javelin throw, I have shown a data to cite the range of values of wind speed that often interact with the velocity of the javelin

Serial No. | Year | Wind Speed (m/s) |
---|---|---|

1 | 2008 | 4.1 |

2 | 2013 | 5.0 |

3 | 2013 | 1.8 |

4 | 1990 | 4.0 |

5 | 1979 | 3.5 |

6 | 1989 | 2.6 |

7 | 1995 | 1.2 |

8 | 1991 | 2.9 |

9 | 1991 | 0.3 |

10 | 1991 | 4.4 |

11 | 2009 | 4.0 |

12 | 2009 | 4.1 |

From the above-mentioned data, we can see that, wind speed that affects the game ranges up to 5 m/sec. Thus, in this IA, we will consider a range of wind speed from null to 5 m/sec in both the directions that of the horizontal direction of the javelin throw.

**Wind is flowing in the direction as that of the javelin:**

The match which I have seen, the velocity of projection of the athlete was 30 m/s. As, wind speed is a variable in this case study, we will take a constant value of the angle of projection. As we have studied that in the previous section that range will be maximum for angle of projection equal to 45°, the angle of projection will be made constant in this section of the IA and equal to 45°.

Therefore, we can compute the value and establish an equation of range as follows:

\(R = \frac{u^2×sin \, \,sin\,2\theta}{g}\)

\(=> R_{max} = \frac{u^2×sinsin\,(2×45)°}{g}=\frac{u^2×1}{g}=\frac{u^2}{g}\)

Here, u is the combined velocity of javelin and wind. From equation (2), we can write,

u = u_{i} + u_{w}

=> u = 30 + u_{w}

Serial No. | Wind Speed (in m/s) | Effective Velocity (in m/s) | Range (in m) |
---|---|---|---|

1 | 0.5 | 30.5 | 94.82671 |

2 | 1 | 31 | 97.96126 |

3 | 1.5 | 31.5 | 101.1468 |

4 | 2 | 32 | 104.3833 |

5 | 2.5 | 32.5 | 107.6707 |

6 | 3 | 33 | 111.0092 |

7 | 3.5 | 33.5 | 114.3986 |

8 | 4 | 34 | 117.8389 |

9 | 4.5 | 34.5 | 121.3303 |

10 | 5 | 35 | 124.8726 |

Sample Calculation

Effective Velocity = (30 + 0.5) m/sec = 30.5 m/sec

\(\text{Range} = \frac{30.5^2}{9.891}=\frac{930.25}{9.81} = 94.82 m\)

**Wind is flowing in the direction opposite to that of the javelin**

The match which I have seen, the velocity of projection of the athlete was 30 m/s. As, wind speed is a variable in this case study, we will take a constant value of the angle of projection. As we have studied that in the previous section that range will be maximum for angle of projection equal to 45°, the angle of projection will be made constant in this section of the IA and equal to 45°.

Therefore, we can compute the value and establish an equation of range as follows:

\(R=\frac{u^2×sinsin\,2\theta}{g}\)

\(=> R_{max }= \frac{u^2×sinsin(2×45)°}{g}=\frac{u^2×1}{g}=\frac{u^2}{g}\)

Here, u is the combined velocity of javelin and wind. From equation (2), we can write,

\(u=\biggl{|}u_i-u_w\biggl{|}\)

=> u = 10 - u_{w}

Serial No. | Wind Speed (in m/s) | Effective Velocity (in m/s) | Range (in m) |
---|---|---|---|

1 | 0.5 | 29.5 | 88.7105 |

2 | 1 | 29 | 85.72885 |

3 | 1.5 | 28.5 | 82.79817 |

4 | 2 | 28 | 79.91845 |

5 | 2.5 | 27.5 | 77.0897 |

6 | 3 | 27 | 74.31193 |

7 | 3.5 | 26.5 | 71.58512 |

8 | 4 | 26 | 68.90928 |

9 | 4.5 | 25.5 | 66.2844 |

10 | 5 | 25 | 63.7105 |

Sample Calculation:

Effective Velocity = (30 - 0.5) m/sec = 29.5 m/sec

\(Range = \frac{29.5^2}{9.81}=\frac{870.25}{9.81}=88.71\, m\)

**Investigation on trajectory of Javelin:**

We know that from equation 4 and 5,

\(S_x=u_x\,t+\frac{1}{2}a_xt^2\)

=> *s _{x} = u cos cos*

*=> x = u cos cos **θ** × t*

\(S_y=u_y\,t+\frac{1}{2}a_yt^2\)

\( => y = u \ sin\ sin \theta ×t - \frac{1}{2}gt^2\)

\( => y = u \, \, sin \, \, sin \theta × \frac{x}{ucoscos\,\theta}-\frac{1}{2}g(\frac{x}{ucoscos\,\theta})^2\)

\(=> y = \frac{x}{tantan\,\theta}-\frac{1}{2}g(\frac{x}{ucoscos\,\theta})^2\)

**Investigation of Air Resistance on Javelin:**

We know that,

*v = u + at*

*v _{y }= u_{y }- gt*

*=> v _{y }= u sin sin θ - gt*

Therefore, Air Resistance in Y- Direction:

In general case, the values of the different parameters are as follows:

\(\rho= 0.45Kg.m^{-3}\biggl|C_D = 0.025\biggl|\)

Radius of cross section of javelin = 0.0238 m

Area of cross section of javelin = \(\pi\)r^{2} = 3. 414 × 0. 023^{2} = 0. 00179 m^{2}

^{\(F=\frac{\rho C_DA}{2}×v^2\)}

^{\(=>F= \frac{(0.45)(0.025)(0.00179)}{2}×v^2_y\)}

\(=> F = 0. 000402 × v^2_y\)

Let us consider that the velocity of projection is 30 m/sec and the angle of projection is 45°, then the air resistance will vary as shown below:

=> *F* = 0. 000402 × (*u sin sin* *θ* - *gt*)^{2}

=> *F* = 0. 000402 × (30 *sin sin* 45 - 9.81*t*)^{2}

=> *F* = 0. 1809 - 0. 167*t* + 0. 0386*t*^{2}

Again,

*v = u + at*

*v _{x} = u_{x}*

*=> v _{x} = u cos cos θ*

Therefore, Air Resistance in X- Direction:

\(F=\frac{\rho C_DA}{2} × v^2\)

\(=> F = \frac{(0.45)(0.025)(0.00179)}{2}×v^2_x\)

\(=> F = 0. 000402 × v^2_x\)

Let us consider that the velocity of projection is 30 m/sec and the angle of projection is 45°, then the air resistance will vary as shown below:

=> *F* = 0. 000402 × (*u cos cos** **θ*)^{2}

=> *F* = 0. 000402 × (30 *cos cos* 45)^{2}

=> *F* = 0. 1809 *N*

Mass of Javelin = 0.081 Kg^{10}

Negative acceleration in X Axis = \(\frac{0.1809}{0.015} = 2. 21 m/s^2\)

Therefore, final velocity in X Axis = *u cos cos θ* - 2. 21*t*

Negative acceleration in Y Axis = \(\frac{0.1809-0.167t+0.0386t^2}{0.0815}= (2. 21 - 2. 04t + 0. 473t^2) m/s^2\)

Therefore, final velocity in Y Axis = *u sin sin **θ* - (2. 21 - 2. 04t + 0. 473t^{2} + *g*)*t*

Trajectory of Javelin with Air resistance

We know that from equation 4 and 5,

\(S_x = u_xt+\frac{1}{2}a_xt^2\)

\(=> sx = u\ cos\ cos \ \theta ×t -\frac{1}{2}(2.21)t^2\)

=> *x = u cos cos θ* ×*t* - 1. 105*t*^{2}…(10)

\(S_y = u_yt+\frac{1}{2}a_yt^2\)

\(=> y = u\ \ sin\ \ sin \theta ×t -\frac{1}{2}\biggl(2. 21 - 2. 04t + 0. 473t^2 + g\biggl)t^2\) …(11)

Here we will consider three angles of projection.

In first case, angle of projection is equal to 30° and initial velocity is 30 m/sec.

From equation 10, we can write

*x* = 30 × *cos cos* 30° × *t* - 1. 105*t*^{2}

\(=> x = 15 \sqrt{3t}-1.105t^2\)

\(=> x = 15t^2 - {15}\sqrt{3t}+x=0\)

\(=> t = \frac{15\sqrt{3}±\sqrt{675-4.42x}}{2.21}\)

From the above-mentioned equation of time t, it is clear that the equation will not have a negative sign because if x is zero then then value of t will become zero. If the sign is negative then the value of time will become negative which is absurd to get. So, the equation of time will be as follows:

\(∴ t = \frac{15\sqrt{3}+\sqrt{675-4.42x}}{2.21}\)

Now, from equation 11, we can write,

\(y = 30 × sin sin 30° ×t -\frac{1}{2}\biggl(2. 21 - 2. 04t + 0. 473t^2 + g\biggl)t^2\)

\(=> y = 15t - \biggl(1. 105 - 1. 02t + 0. 2365t^2 + 4. 9\biggl)t^2\)

=> *y* = 15*t* - 1. 105*t*^{2 }+ 1. 02*t*^{3 }- 0. 2365*t*^{4 }- 4.9*t*^{2}

=> y = 15t - 6t^{2 }+ 1. 02t^{3 }- 0. 2365t^{4}

\(=> y = 15× \frac{15\sqrt{3+}\sqrt{675-4.42x}}{2.21}-6(\frac{15\sqrt{3+}\sqrt{675-4.42x}}{2.21})^2 + 1.02\)

\((\frac{15\sqrt{3+}\sqrt{675-4.42x}}{2.21})^3- 0. 2365 (\frac{15\sqrt{3+}\sqrt{675-4.42x}}{2.21})^4\)

This is the equation of trajectory of javelin with air resistance for angle of projection equal to 30°.

In second case, angle of projection is equal to 45° and initial velocity is 30 m/sec.

From equation 10, we can write

*x* = 30 ×* cos cos* 45 × *t* - 1. 105*t*^{2}

\(=> x = {15}\sqrt{2t}-1.105t^2\)

\(=> 1. 105t^2 - {15}\sqrt{2t}+x=0\)

\(=> t = \frac{15\sqrt{2±}\sqrt{450-4.42x}}{2.21}\)

From the above-mentioned equation of time t, it is clear that the equation will not have a negative sign because if x is zero then then value of t will become zero. If the sign is negative then the value of time will become negative which is absurd to get. So, the equation of time will be as follows

\(∴ t =\frac{15\sqrt{2}+\sqrt{450-4.42x}}{2.21}\)

Now, from equation 11, we can write,

\(y = 30 × \, \,sin \, \,sin 45° × t - \frac{1}{2}\bigg(2. 21 - 2. 04t + 0. 473t2 + g\biggl)t^2\)

\(=> y = {15}\sqrt{2t}-\biggl(1. 105 - 1. 02t + 0. 2365t^2 + 4. 9\biggl)t^2\)

\(=> y = {15}\sqrt{2t} - 1.105t^2 + 1.02t^3 - 0.2365t^4 - 4.9t^2\)

\(=> y = {15}\sqrt{2t} - 6t^2 + 1.02t^3 - 0.2365t^4\)

\(=> y = 15\sqrt{2}×\frac{15\sqrt{2+}\sqrt{450-4.42x}}{2.21}-6(\frac{15\sqrt{2+}\sqrt{450-4.42x}}{2.21})^2\)

\(+ 1.02(\frac{15\sqrt{2+}\sqrt{450-4.42x}}{2.21})^3- 0. 2365(\frac{15\sqrt{2+}\sqrt{450-4.42x}}{2.21})^4\)

This is the equation of trajectory of javelin with air resistance for angle of projection equal to 45°.

In third case, angle of projection is equal to 60° and initial velocity is 30 m/sec.

From equation 10, we can write

*x* = 30 × *cos* 60° × *t* - 1. 105*t*^{2}

=> *x* = 15*t* - 1. 105*t*^{2}

=> 1. 105*t*^{2 }- 15*t* + *x* = 0

\(=> t = \frac{15±\sqrt{225-4.42x}}{2.21}\)

From the above-mentioned equation of time t, it is clear that the equation will not have a negative sign because if x is zero then then value of t will become zero. If the sign is negative then the value of time will become negative which is absurd to get. So, the equation of time will be as follows:

\( ∴ t =\frac{15+\sqrt{225-4.42x}}{2.21}\)

Now, from equation 11, we can write,

\(y = 30 × \,sin \ 60° ×t - \frac{1}{2}\biggl(2. 21 - 2. 04t + 0. 473t^2 + g\biggl)\)

\(=> y ={15}\sqrt{3t}-\biggl(1.105 - 1.02t + 0.2365t^2 + 4.9t^2\biggl)t^2\)

\(=> y = {15}\sqrt{3t}-6t^2 + 1. 02t^3 - 0. 2365t^4\)

\(=> y = {15}\sqrt{3}×\frac{15+\sqrt{225-4.42x}}{2.21}-6(\frac{15+\sqrt{225-4.42x}}{2.21})^2\)

\(1.02(\frac{15+\sqrt{225-4.42x}}{2.21})^3 - 0. 2365 (\frac{15+\sqrt{225-4.42x}}{2.21})^4\)

This is the equation of trajectory of javelin with air resistance for angle of projection equal to 60°.

**Time Period and Range of Javelin Throw for different projection with air resistance:**

In all the cases, the initial velocity of projection will be 30 m/sec.

In first case,

Angle of projection is 30°.

From equation 11, we can write,

\(y = u \ \ sin \ \sin \theta ×t - \frac{1}{2}\biggl(2.21 - 2. 04t + 0. 473t^2 + g\biggl)t^2\)

\(y = 30 × sin\ \ sin 30° ×t - \frac{1}{2}\biggl(2.21 - 2. 04t + 0. 473t^2 + g\biggl)\)

\(=> y = 15t - \biggl(1. 105 - 1. 02t + 0. 2365t^2 + 4. 9\biggl)t^2\)

=> y = 15t - 1. 105t^{2 }+ 1. 02t^{3 }- 0. 2365t^{4 }- 4. 9t^{2}

=> y = 15 t - 6t^{2 }+ 1. 02t^{3 }- 0. 2365t^{4}

In order to find the total time of flight, we have to find the value of t for y = 0:

0. 2365t^{4 }- 1. 02t^{3 }+ 6t^{2 }- 15t = 0

Quartic equation

ax^{4 }+ bx^{3} + cx^{2} + dx + e = 0

Therefore, the time period is 2.96 sec. The value of time cannot be complex in nature thus we will ignore the complex roots of the equation. One root of t is zero because at time t = 0 the displacement of javelin in y direction was also zero (initial position).

Again,

From equation 10, we can write

x = 30 × cos cos 30° × t - 1. 105t^{2}

\(=>x= 15\sqrt{3t}-1.105t^2\)

\(=>x= 15\sqrt{3}×2.96 - 1.105(2.96)^2\)

=> x = 67. 22 metre

In second case,

Angle of projection is 45°.

From equation 11, we can write,

\(y=u \,sin \,sin \,\theta ×t-\frac{1}{2}\biggl(2.21-2.04t+0.473t^2+g\biggl)t^2\)

\(y=30× \,sin \,sin \,45° ×t-\frac{1}{2}\biggl(2.21-2.04t+0.473t^2+g\biggl)t^2\)

\(=>y=15\sqrt{2t}-\biggl(1.105-1.02t+0.2365t^2+4.9\biggl)t^2\)

\(=>y=15\sqrt{2t}-1.\,105t^2+1.\,02t^3-0.\,2365t^4-4.\,9t^2\)

\(=>y=15\sqrt{2t}-6t^2+1.02t^3-0.2365t^4\)

In order to find the total time of flight, we have to find the value of t for y = 0:

0. 2365*t*^{4 }- 1. 02*t*^{3 }+ 6*t*^{2 }- 21. 21*t* = 0

Quartic equation

ax^{4} + br^{3} + cx^{2} + dx + e = 0

Therefore, the time period is 3.81 sec. The value of time cannot be complex in nature thus we will ignore the complex roots of the equation. One root of t is zero because at time t = 0 the displacement of javelin in y direction was also zero (initial position).

Again,

From equation 10, we can write

*x *= 30 × cos cos 45 × t - 1. 105t^{2}

\(=> x = 15\sqrt{2t}-1.105t^2\)

\(=> x = 15\sqrt{2}×3.81 - 1.105(3.81)^2\)

=>* x* = 64. 78 metre

In third case,

Angle of projection is 60°.

From equation 11, we can write,

\(y=u\,sin\,sin\, \theta ×t-\frac{1}{2}\biggl(2.21-2.04t + 0.473t^2+g\biggl)t^2\)

\(y=30×\,sin\,sin\, 60° ×t-\frac{1}{2}\biggl(2.21-2.04t + 0.473t^2+g\biggl)t^2\)

\(=>y=15\sqrt{3t}-\biggl(1.105-1.02t+0.2365t^2+4.9\biggl)t^2\)

\(=>y=15\sqrt{3t}-1.105t^2+1.02t^3-0.2365t^4-4.9t^2\)

\(=>y={15}\sqrt{3t}-6t^2+1.02t^3-0.2365t^4\)

In order to find the total time of flight, we have to find the value of t for y = 0:

0. 2365*t*^{4 }- 1. 02*t*^{3 }+ 6*t*^{2 }- 25. 98*t* = 0

Quartic equation

*ax ^{4} + bx*

Therefore, the time period is 4.32 sec. The value of time cannot be complex in nature thus we will ignore the complex roots of the equation. One root of t is zero because at time t = 0 the displacement of javelin in y direction was also zero (initial position).

Again,

From equation 10, we can write

*x* = 30 × *cos cos *60° ×* t* - 1. 105*t*^{2}

=>* x* = 15t - 1. 105*t*^{2}

=> *x* = 15 × 4. 32 - 1. 105(4.32 )^{2}

=> *x* = 44. 17 metre