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]

The 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.

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

This part of the problem introduces an incline to the equation, changing the angle required for the optimal range, as the projectile now should not necessarily be thrown at 45° to achieve the maximum range. As highlighted on the diagram, the projectile is being thrown onto the incline. The greatest distance along the incline leads to the greatest range along the 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
tanθ
and therefore the equation of the plane can be expressed as

 
y = tanγ x
, with
γ
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 = vcos(θ))t
t=xvcosθ

This equation for t can be substituted into the equation of motion for the y value,
y=ut+12at2
, leading to:
y=vsin⁡θxvcosθ–gx22v2cos2θ

This can then be equated to the equation for the slope
y=tan⁡γx
giving the equation:
tan⁡γx=vsin⁡θxvcosθ–gx22v2cos2θ

and then dividing both sides by x (ignoring the trivial case when
x=0
) gives the equation 
tan⁡γ=vsin⁡θvcos⁡θ–gx2v2cos2θ
tan⁡γ=tan⁡θ–gx2v2cos2θ

Then rearranging to find x
tanγ–tanθ2v2cos2θ=–xg
x=–v2g2cos2θtanγ–sin2θ

This can then be differentiated with respect to
θ
in order to find the maximum distance and the angle that achieves this distance.
dxdθ–v2g2cos2θtanγ–sin2θ
=–v2g4cosθsinθtanγ–2cos2θ
=–v2g2sin2θtanγ–2cos2θ

Setting this to 0 to find the optimum angle
–v2g2sin2θtanγ–2cos2θ=0

The trivial case where
v=0 
can be ignored, therefore both sides can be divided by
–v2g.
–2cos2θ=2sin2θtanγ
–cos2θsin2θ=tanγ

tan2θ=–1tanγ

This can be rewritten as
2θ=arctan–cotγ

As
tan(x)
is an odd function,
–fx=f–x
and
cotx
=
tanπ2 – x
2θ= arctan⁡(γ–π2)

In order to find the positive value which is required, pi must be added to go from the 4th quadrant to the 2nd quadrant.
2θ=arctantan⁡γ – π2+π
2θ=γ+π2
θ=γ2+π4

This result can be validated as when the angle
γ= 0,
the equation becomes that
θ =π4+02
leading to
θ=45∘
, which is true as shown by the previous derivation in section 1.

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=vt+12at2.
The new equation can be expressed as
y–h=vsinθt–12
y=h+vsinθt–12

By substituting in the values, there is a quadratic equation for the value of time t
y=–12gt2+vsinθt+h

Using the Quadratic Formula,
–b±b2–4ac2a
, gives t as
t=–vsinθ±v2sin2θ+2gh–g

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=vcosθt
giving the equation:
x=v2sinθcosθ+vcosθv2sin2θ+2ghg

This equation can be differentiated in order to find the maximum range.
dxdθv2sinθcosθ+vcosθv2sin2θ+2ghg
=1gv2cos2θ–sin2θ–vsinθv2sin2θ+2gh+2v3sinθcos2θ2v2sin2θ+2gh

This can then be equated to 0 in order to find the optimum value of
θ.
v2cos2θ–sin2θ=vsinθv2sin2θ+2gh–2v3sinθcos2θ2v2sin2θ+2gh

Squaring both sides to remove the square roots.
(v^4(cos⁡^4(θ)–2sin⁡^2(θ)cos⁡^2(θ))+sin⁡^4(θ))
=v^2sin^2(θ)(v^2sin^2(θ)+2gh)–2v^4sin^2(θ)cos^2(θ)+v^6sin^2(θ)cos^4(θ)v^2sin^2(θ)+2gh

v4cos4θ–v42sin2θcos2θ+v4sin4θ=
v4sin4θ+v22ghsin2θ–v42sin2θcos2θ+v6sin2θcos4θv2sin2θ+2gh

Divide both sides by
v2
as the trivial case where
v=0
can be ignored.
v2cos4θ=2ghsin2θ+v4sin2θcos4θv2sin2θ+2gh

 
v4sin2θcos4θ+v2cos4θ2gh=v22ghsin4θ+2gh2sin2θ+v4sin2θcos4θ

Then divide both sides by
2gh
as for this section of the essay, the height is above 0.
v2cos4θ=v22ghsin2θ+sin4θ

Using the trigonometric identities
cos4θ
can be written as
1–sin2θ2
v^2–v^2 2sin^2(θ)+v^2 sin^4(θ))= 2ghsin^2(θ)+v^2sin^4(θ)

 
v2–2v2sin2θ=2ghsin2θ
v22v2+2gh=sin2θ
±v2v2+2gh=sinθ

For the range of the projectile in the x direction, the angle must be between 0 and
π2
. The negative value of the above equation can be ignored as
sinθ
is always positive between 0 and
π
.

Therefore, the optimal angle of a projectile from a non-zero height is:
arcsinv2v2+2gh

As h approaches 0, the value for the optimal angle becomes
arcsin12
which is 45° or
π4
radians, validating the solution when h = 0, which is section 1.

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
θ=γ2+π4
, with
γ
being the angle of the inclined plane. The optimal angle for the maximum distance from a non-zero height is
sin–1⁡v2v2+2gh
. 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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[8] Buchwald, J & Fox, R (2013)

[12] Buchwald, J & Fox, R (2013)

[13] Serway, R. et al (2009)