Optimal Angles of Projectile Motion
✅ Paper Type: Free Essay | ✅ Subject: Physics |
✅ Wordcount: 7915 words | ✅ Published: 23rd Sep 2019 |
Optimal Angles of Projectile Motion
What is the optimal angle of a projectile from a given height?
Contents
1.1 Brief History of the Problem
1.2 Deriving the Equations of Motion
1. Introduction
Projectile motion is the motion experienced by a particle or object with the only force acting upon the projectile being gravity.[1] Aristotle, the Ancient Greek mathematician and philosopher, first investigated into the motion of objects which was then expanded upon and corrected by Galileo Galilei using experimental methods, before being mathematically shown by Newton, using calculus.[2] Projectile motion plays a crucial role in many aspects of human life, such as sport, the military and even space exploration. Understanding the fundamentals of projectile motion can lead to a greater understanding of not only our world but outside as well, with Newton using the fundamentals of motion and his laws of motion and gravity to be able to track planetary motion, using work investigated by Johannes Kepler. Kepler tracked planetary motion without the understanding of the forces between them, and his work, helped people gain a greater understanding of the universe, using the fundamentals of motion to track the elliptical paths of planets.[3]
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Essay Writing ServiceThe angle that the projectile is thrown at is one of the variables which can be easily changed and greatly affects the optimal range of the projectile. After investigating the optimal angle as projected from the ground on a flat plane, the next thing to be investigated is the motion on an inclined plane with the new optimal angle for distance being dependent on the angle of the incline. After this, the optimal angle as projected from a height is important to be investigated, as in many real life circumstances, the object, such as ball thrown in sports, is not projected from the ground, but from a height, and therefore the range, and optimal angle for the furthest range would differ, and can be much more useful than the optimal angle from a zero height.
The main method used throughout this extended essay is finding the range along the x axis, by using the equations of motion and combining the y direction and x direction by the variable they share in common, time. After finding a range based on the variables used in the equations of motion, the range can be expressed and then differentiated, with respect to θ, setting this differentiated equation equal to 0. By solving the resulting equation for the angle of then the optimal angle can be found, as this would be the maximum point of the velocity equation, and thus the maximum distance along the x axis.
One of the most crucial variables in all of kinematics is the acceleration due to gravity, denoted by g. The quantity of this value on the surface of the Earth was first discovered by Henry Cavendish, an 18th Century Physicist, who, using a torsion pendulum, measured the value of g, leading to the value of approximately 9.8 ms^{-2}.[4]
1.1 Brief History of the Problem
The topic of kinematics and the effects upon objects in motion was first investigated by the Ancient Greeks, mainly Aristotle who viewed motion as being in one of two states, either violent motion or natural motion[5], stating that motion is a change from potentiality to actuality.[6] Aristotle’s explanation of the accelerating of bodies due to gravity was that it depended on the “heaviness” of the body thatis falling, and that a body is attracted to its natural place by its heaviness.[7] Over 1000 years after Aristotle, Galileo Galilei, who started his scientific career by annotating the work of Aristotle, developed the equations of motions by publicly experimenting on the fall of heavy bodies , by showing that two “moveables” do not move with speeds proportional to their weight, as stated by Aristotle, but all move at the same speed,[8] deducing the equation $$
u = the initial velocity of a projectile(ms^{-1})
v = the final velocity of the projectile(ms^{-1})
t = the time passed since the projectile has been thrown (seconds)
s = the displacement of the projectile (metres) from initial to the final position of the projectile
Acceleration is defined as the rate of change of velocity with respect to time. This can be expressed as the differential equation: $$θ = the angle that the projectile is thrown at from the ground
v = the velocity of the projectile
When an object’s speed is increasing, the acceleration is positive in the direction of the vector. When the object’s speed is decreasing, the acceleration is therefore negative. The horizontal acceleration is 0 as the projectile does not accelerate in the x direction. The acceleration is -g in the vertical direction, as it is constantly accelerating down due to gravity. As gravity is being enacted downwards toward the Earth, the acceleration is negative.
The diagram is used to show the method for solving the range and angle for 2D projectile motion, investigating the motion in each direction. Projectile motion can be described as the superposition of two independent forces in the x and y direction,[13] with the variable joining the two directions being time, meaning that the time can be derived from the y direction and then be substituted into the x direction in order to find the range and optimise it. The initial y component of velocity is $$x axis, as the two are proportional. The gradient of the incline needs to be considered when calculating the optimal angle. As the gradient is expressed as “rise over run”, the gradient of the incline is $\mathit{tan}\left(\theta \right)$
and therefore the equation of the plane can be expressed as
$\mathit{y}=\mathit{tan}\left(\gamma \right)x$
, with $\gamma $
being equal to the angle of the inclined plane. This equation for the gradient of an incline can then be equated to an equation for y, to rearrange for x and find the maximum range.
First, the time must be expressed using the equations of motion, in respect to the x direction.
$x\mathrm{}=\mathrm{}\mathit{vcos}\left(\theta \right))t$
$t=\frac{x}{\mathit{vcos}\left(\theta \right)}$
This equation for t can be substituted into the equation of motion for the y value, $y=\mathit{ut}+\frac{1}{2}a{t}^{2}$
, leading to:
$y=v\mathrm{sin}\left(\theta \right)\frac{x}{\mathit{vcos}\left(\theta \right)}\u2013\frac{g{x}^{2}}{2{v}^{2}\mathit{co}{s}^{2}\left(\theta \right)}$
This can then be equated to the equation for the slope $y=\mathrm{tan}\left(\gamma \right)x$
giving the equation:
$\mathrm{tan}\left(\gamma \right)x=v\mathrm{sin}\left(\theta \right)\frac{x}{\mathit{vcos}\left(\theta \right)}\u2013\frac{g{x}^{2}}{2{v}^{2}\mathit{co}{s}^{2}\left(\theta \right)}$
and then dividing both sides by x (ignoring the trivial case when $x=0$
) gives the equation
$\mathrm{tan}\left(\gamma \right)=\frac{v\mathrm{sin}\left(\theta \right)}{v\mathrm{cos}\left(\theta \right)}\u2013\frac{\mathit{gx}}{2{v}^{2}\mathit{co}{s}^{2}\left(\theta \right)}$
$\mathrm{tan}\left(\gamma \right)=\mathrm{tan}\left(\theta \right)\u2013\frac{\mathit{gx}}{2{v}^{2}\mathit{co}{s}^{2}\left(\theta \right)}$
Then rearranging to find x
$\left(\mathit{tan}\left(\gamma \right)\u2013\mathit{tan}\left(\theta \right)\right)\left(2{v}^{2}\mathit{co}{s}^{2}\left(\theta \right)\right)=\u2013\mathit{xg}$
$x=\u2013\frac{{v}^{2}}{g}\left(2\mathit{co}{s}^{2}\left(\theta \right)\mathit{tan}\left(\gamma \right)\u2013\mathit{sin}\left(2\theta \right)\right)$
This can then be differentiated with respect to $\theta $
in order to find the maximum distance and the angle that achieves this distance.
$\frac{\mathit{dx}}{\mathit{d\theta}}\left(\u2013\frac{{v}^{2}}{g}\left(2\mathit{co}{s}^{2}\left(\theta \right)\mathit{tan}\left(\gamma \right)\u2013\mathit{sin}\left(2\theta \right)\right)\right)$
$=\u2013\frac{{v}^{2}}{g}\left(4\mathit{cos\theta sin}\left(\theta \right)\mathit{tan}\left(\gamma \right)\u20132\mathit{cos}\left(2\theta \right)\right)$
$=\u2013\frac{{v}^{2}}{g}\left(2\mathit{sin}\left(2\theta \right)\mathit{tan}\left(\gamma \right)\u20132\mathit{cos}\left(2\theta \right)\right)$
Setting this to 0 to find the optimum angle
$\u2013\frac{{v}^{2}}{g}\left(2\mathit{sin}\left(2\theta \right)\mathit{tan}\left(\gamma \right)\u20132\mathit{cos}\left(2\theta \right)\right)=0$
The trivial case where $v=0$
can be ignored, therefore both sides can be divided by $\u2013\frac{{v}^{2}}{g}.$
$\u20132\mathit{cos}\left(2\theta \right)=2\mathit{sin}\left(2\theta \right)\mathit{tan}\left(\gamma \right)$
$\u2013\frac{\mathit{cos}\left(2\theta \right)}{\mathit{sin}\left(2\theta \right)}=\mathit{tan}\left(\gamma \right)$
$\mathit{tan}\left(2\theta \right)=\u2013\frac{1}{\mathit{tan}\left(\gamma \right)}$
This can be rewritten as
$2\theta =\mathit{arctan}\left(\u2013\mathit{cot}\left(\gamma \right)\right)$
As $\mathit{tan}\left(x\right)$
is an odd function, $\u2013f\left(x\right)=f\left(\u2013x\right)$
and $\mathit{cot}\left(x\right)$
= $\mathit{tan}\left(\frac{\pi}{2}\u2013\mathit{x}\right)$
$2\theta =\mathrm{}\mathit{arctan}(\gamma \u2013\frac{\pi}{2})$
In order to find the positive value which is required, pi must be added to go from the 4th quadrant to the 2nd quadrant.
$2\theta =\mathit{arctan}\left(\mathrm{tan}\left(\mathit{\gamma}\u2013\frac{\pi}{2}+\pi \right)\right)$
$2\theta =\gamma +\frac{\pi}{2}$
$\theta =\frac{\gamma}{2}+\frac{\pi}{4}$
This result can be validated as when the angle $\mathrm{\gamma}=0,$
the equation becomes that
$\mathit{\theta \u2008}=\frac{\pi}{4}+\frac{0}{2}$
leading to $\theta ={45}^{\circ}$
, which is true as shown by the previous derivation in section 1.
4.Developing the Final Solution
The next part of the problem is to investigate how the optimal angle changes depending on throwing the projectile from some height, h, above the ground. This part of the investigation is most relevant to real life application, as it is unlikely that a projectile will be thrown from the ground, but instead is more likely to be thrown from a height, e.g. throwing a ball from shoulder-height in different sports.
To first investigate this problem, a new equation must be made from the equation of motion previously used, $y=\mathit{vt}+\frac{1}{2}a{t}^{2}.$
The new equation can be expressed as
$y\u2013h=\mathit{vsin}\left(\theta \right)t\u2013\frac{1}{2}$
$y=h+\mathit{vsin}\left(\theta \right)t\u2013\frac{1}{2}$
By substituting in the values, there is a quadratic equation for the value of time t
$y=\u2013\frac{1}{2}g{t}^{2}+\mathit{vsin}\left(\theta \right)t+h$
Using the Quadratic Formula, $\frac{\u2013b\pm \sqrt{{b}^{2}\u20134\mathit{ac}}}{2a}$
, gives t as
$t=\frac{\u2013\mathit{vsin}\left(\theta \right)\pm \sqrt{{v}^{2}\mathit{si}{n}^{2}\left(\theta \right)+2\mathit{gh}}}{\u2013g}$
There is an interpretation for the negative value of the square root, and thus the negative value for the time. The negative time represents the motion of the projectile before it has been projected as highlighted in this problem. However, even though this is a valid interpretation of the time value, it is not useful for finding the optimum angle for distance, and thus only the positive value for the square root will be used henceforth. This time can then be put into the equation of motion $x=\mathit{vcos}\left(\theta \right)t$
giving the equation:
$x=\frac{{v}^{2}\mathit{sin}\left(\theta \right)\mathit{cos}\left(\theta \right)+\mathit{vcos}\left(\theta \right)\sqrt{{v}^{2}\mathit{si}{n}^{2}\left(\theta \right)+2\mathit{gh}}}{g}$
This equation can be differentiated in order to find the maximum range.
$\frac{\mathit{dx}}{\mathit{d\theta}}\left(\frac{{v}^{2}\mathit{sin}\left(\theta \right)\mathit{cos}\left(\theta \right)+\mathit{vcos}\left(\theta \right)\sqrt{{v}^{2}\mathit{si}{n}^{2}\left(\theta \right)+2\mathit{gh}}}{g}\right)$
$=\frac{1}{g}\left({v}^{2}\left(\mathit{co}{s}^{2}\left(\theta \right)\u2013\mathit{si}{n}^{2}\left(\theta \right)\right)\u2013\mathit{vsin}\left(\theta \right)\sqrt{{v}^{2}\mathit{si}{n}^{2}\left(\theta \right)+2\mathit{gh}}+\frac{2{v}^{3}\mathit{sin}\left(\theta \right)\mathit{co}{s}^{2}\left(\theta \right)}{2\sqrt{{v}^{2}\mathit{si}{n}^{2}\left(\theta \right)+2\mathit{gh}}}\right)$
This can then be equated to 0 in order to find the optimum value of $\theta .$
${v}^{2}\left(\mathit{co}{s}^{2}\left(\theta \right)\u2013\mathit{si}{n}^{2}\left(\theta \right)\right)=\mathit{vsin}\left(\theta \right)\sqrt{{v}^{2}\mathit{si}{n}^{2}\left(\theta \right)+2\mathit{gh}}\u2013\frac{2{v}^{3}\mathit{sin}\left(\theta \right)\mathit{co}{s}^{2}\left(\theta \right)}{2\sqrt{{v}^{2}\mathit{si}{n}^{2}\left(\theta \right)+2\mathit{gh}}}$
Squaring both sides to remove the square roots.
$(v^4(\mathrm{cos}^4\left(\theta \right)\u20132\mathrm{sin}^2\left(\theta \right)\mathrm{cos}^2\left(\theta \right))+\mathrm{sin}^4(\theta \left)\right)$
$=v^2\mathit{sin}^2\left(\theta \right)(v^2\mathit{sin}^2(\theta )+2\mathit{gh})\u20132v^4\mathit{sin}^2\left(\theta \right)\mathit{cos}^2\left(\theta \right)+\frac{v^6\mathit{sin}^2\left(\theta \right)\mathit{cos}^4\left(\theta \right)}{v^2\mathit{sin}^2\left(\theta \right)+2\mathit{gh}}$
${v}^{4}\mathit{co}{s}^{4}\left(\theta \right)\u2013{v}^{4}2\mathit{si}{n}^{2}\left(\theta \right)\mathit{co}{s}^{2}\left(\theta \right)+{v}^{4}\mathit{si}{n}^{4}\left(\theta \right)=$
${v}^{4}\mathit{si}{n}^{4}\left(\theta \right)+{v}^{2}2\mathit{ghsi}{n}^{2}\left(\theta \right)\u2013{v}^{4}2\mathit{si}{n}^{2}\left(\theta \right)\mathit{co}{s}^{2}\left(\theta \right)+\frac{{v}^{6}\mathit{si}{n}^{2}\left(\theta \right)\mathit{co}{s}^{4}\left(\theta \right)}{{v}^{2}\mathit{si}{n}^{2}\left(\theta \right)+2\mathit{gh}}$
Divide both sides by ${v}^{2}$
as the trivial case where $v=0$
can be ignored.
${v}^{2}\mathit{co}{s}^{4}\left(\theta \right)=2\mathit{ghsi}{n}^{2}\left(\theta \right)+\frac{{v}^{4}\mathit{si}{n}^{2}\left(\theta \right)\mathit{co}{s}^{4}\left(\theta \right)}{{v}^{2}\mathit{si}{n}^{2}\left(\theta \right)+2\mathit{gh}}$
${v}^{4}\mathit{si}{n}^{2}\left(\theta \right)\mathit{co}{s}^{4}\left(\theta \right)+{v}^{2}\mathit{co}{s}^{4}\left(\theta \right)2\mathit{gh}={v}^{2}2\mathit{ghsi}{n}^{4}\left(\theta \right)+{\left(2\mathit{gh}\right)}^{2}\mathit{si}{n}^{2}\left(\theta \right)+{v}^{4}\mathit{si}{n}^{2}\left(\theta \right)\mathit{co}{s}^{4}\left(\theta \right)$
Then divide both sides by $2\mathit{gh}$
as for this section of the essay, the height is above 0.
${v}^{2}\mathit{co}{s}^{4}\left(\theta \right)={v}^{2}2\mathit{ghsi}{n}^{2}\left(\theta \right)+\mathit{si}{n}^{4}\left(\theta \right)$
Using the trigonometric identities $\mathit{co}{s}^{4}\left(\theta \right)$
can be written as ${\left(1\u2013\mathit{si}{n}^{2}\left(\theta \right)\right)}^{2}$
$v^2\u2013v^2\mathrm{}2\mathit{sin}^2\left(\theta \right)+v^2\mathrm{}\mathit{sin}^4\left(\theta \right))=\mathrm{}2\mathit{ghsin}^2(\theta )+v^2\mathit{sin}^4(\theta )$
${v}^{2}\u2013{2v}^{2}\mathit{si}{n}^{2}\left(\theta \right)=2\mathit{ghsi}{n}^{2}\left(\theta \right)$
$\frac{{v}^{2}}{2{v}^{2}+2\mathit{gh}}=\mathit{si}{n}^{2}\left(\theta \right)$
$\pm \frac{v}{\sqrt{2{v}^{2}+2\mathit{gh}}}=\mathit{sin}\left(\theta \right)$
For the range of the projectile in the x direction, the angle must be between 0 and $\frac{\pi}{2}$
. The negative value of the above equation can be ignored as $\mathit{sin}\left(\theta \right)$
is always positive between 0 and $\pi $
.
Therefore, the optimal angle of a projectile from a non-zero height is:
$\mathit{arcsin}\left(\frac{v}{\sqrt{2{v}^{2}+2\mathit{gh}}}\right)$
As h approaches 0, the value for the optimal angle becomes $\mathit{arcsin}\left(\frac{1}{\left(\sqrt{2}\right)}\right)$
which is 45^{° }or $\frac{\pi}{4}$
radians, validating the solution when h = 0, which is section 1.
Conclusion
In conclusion, the optimal angle for a projectile, projected from the ground is 45^{°} , and when the projectile is projected up a slope the optimal angle for maximum distance is dependent on the angle of the slope but can be worked out using the equation $\theta =\frac{\gamma}{2}+\frac{\pi}{4}$
, with $\gamma $
being the angle of the inclined plane. The optimal angle for the maximum distance from a non-zero height is ${\mathrm{sin}}^{\u20131}\left(\frac{v}{\sqrt{2{v}^{2}+2\mathit{gh}}}\right)$
. Interestingly, the optimal angle of a projectile which was not projected from the ground is not only dependent on the height, but also the velocity with which the projectile is hurled. This means, for example, that from a height of 1 metre, the optimal angle could change depending on the force with which the projectile has been thrown. For the other two cases, the angle is not dependent at all on the velocity that the projectile is thrown at. These answers are important in real life as they highlight the potential best way to hurl a projectile depending on a variety of factors, Even though air resistance will affect the optimal angle in real life scenarios, for sport and other real life activities, these values, and the understanding of why these values work is important for optimising many parts of where projectile motion plays a key role.
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