# Projectile motion in Physics

A projectile is an object upon which only the force of gravity is acting. For our discussions  we assume that the influence of air resistance can be neglected.

For example, an object dropped from rest at some height is a projectile as long as it is in the air.

Generally, the projectile is an object thrown into the air:

• vertically upward or downward,
• horizontally,
• or at any angle to the Earth surface.

In Physics we will study here the motion of a projectile, its velocity, acceleration, displacement, and  trajectory, that is the path drawn by a moving projectile. As already mentioned we neglect air resistance.

We must be aware that equations presented in this paragraph will be true for projectiles such as a small stone, metal ball from a ball bearing, or a small ball from heavy plastic. A feather or sheet of paper will not move as a projectile because the friction due to presence of air is too large to be neglected.

The free fall motion from Chapter 1. p. 5 is a special case (one dimensional) of a projectile motion. Here we start with a quite general, classical case of a projectile motion. The situation for deriving equations is depicted in Fig. M2.3.

We will describe the motion in the Cartesian coordinate two dimensional system x, y, with y axis being oriented vertically.

The small object, theoretically a point without dimensions, was thrown from the position x=0, y=0 of this coordinate system with the initial velocity , at an angle α from the positive direction of x axis.

The projectile (that is our object) will move along the trajectory drawn by the dotted line in Fig. M2.2. This trajectory is a part of the parabola and this will be evident from equations we will derive here.

To study the motion of this object we decompose the initial velocity  into its Cartesian components v0x and v0y. These components are also vectors, but we omit arrows in this notation because subscript x and y implies values with direction. From the trigonometry it follows that

v0x= v0 cosa            (M2.1)

v0y= v0 sina            (M2.2)

We know from previous paragraphs that:

The motion along the x axis is completely independent from the motion along the y axis.

Along the horizontal x axis there is a motion with constant velocity v0x. The distance traveled along the x axis as a function of time can be calculated using Eq. M1.23 with x0=0, so we can write

x(t) = v0x t        (M2.3)

As a time origin we take t = 0. We assume that the object is moving according to this equation as long as it is above the level of the x axis, which we arbitrarily chose as zero level or ground level. This situation imitates, for example, the motion of a golf ball if it is struck not too hard, because at high speed (strong hit) the air resistance cannot be neglected.

Distance traveled along the y axis is governed by Eq. M1.30, which we rewrite into the form reflecting notations from the Fig. M2.2

(M2.4)

and velocity along the y axis may be found from Eq. M1.28 after writing it as

vy (t) = v0y – g t           (M2.5)

with g being gravitational acceleration.

There are to main questions asked for such a case of projectile motion:

1) What is a largest vertical displacement for a given angle a ,

2) What is a largest horizontal displacement for a given angle a.

The largest vertical displacement is denoted H on Fig. M2.2 and can be calculated with Eq. M.2.4 after substituting into it the time of rising of the projectile – tr, which, in turn can be calculated from Eq. M2.5.

The rising of the projectile continues till the moment its vertical component of velocity decreases to zero, so from Eq. M2.5 we have

0 = v0y –g tr

so

tr = voy / g

or, taking into account Eq. M2.2

tr = v0 sina / g     (M2.6)

Substituting this expression for rising time into Eq. M2.4, describing the vertical displacement, we get after very little algebra, the largest horizontal displacement obtained by the projectile

(M2.7)

notation “sin2α “ is equivalent to “(sinα)2 and can be used alternatively.

At this point we answered question 1) - What is the maximum vertical displacement.

To answer the question 2) we must calculate the total time projectile will be above the level of the x axis. We already calculated the time of rising to the height H  - Eq. M2.6. It is easy to prove (see Problems to this Chapter) that the time of falling tf from this height is equal to the time of rising, that is

tf = tr            (M2.8)

So the total time tT the projectile traveling in the x direction is tT= 2 tr  and, after substituting Eq.M2.6 for tr

tT = 2 v0 sinα / g         (M2.9)

Substituting this expression for time of motion into Eq. M2.3 and taking into account Eq. M2.1 for the horizontal component of initial velocity we will get equation for the largest horizontal displacement, denoted by R on Fig.M2.2

(M2.10)

From trigonometry we know that

2 sinα cosα = sin(2α

so we can write Eq. M2.10 in the form

R = v02 sin(2α) / g         (M2.11)

The next question we can ask is: for which angle of firing a, with a given initial velocity v0, will

a)     the largest vertical displacement have its maximum,

b)     the largest horizontal displacement have its maximum.

We can use calculus for solving these questions. To answer question a) we must find the maximum of the function given by Eq. M2.7. The standard procedure for this is simple. Find the derivative of this function with respect to a, compare it to zero and solve this equation.

Comparing the right hand side of this equation to zero we get the trigonometric equation

(M2.12)

or, in another form

(M2.12a)

which is the condition for the existence of a maximum or minimum of a function. Solving it with respect to α gives

α = 00  and α = 900

The first solution, α = 00 leads to vertical displacement equal to zero – the minimum of function given by Eq. M2.7. Therefore the maximum vertical displacement is obtained for the firing angle α = 900. It is quite obvious, that maximum height is obtained for throwing an object vertically, but now you have the analytically derived proof.

The angle at which the maximum horizontal displacement is obtained can be found similarly.

Now solving the equation

we find

2α = 900,

so

α = 450.

If you are not familiar with calculus, do not worry, simply remember the final results. At a given initial firing velocity of the projectile:

maximum vertical displacement is obtained when firing at an angle of 900,

and

maximum horizontal displacement is obtained when firing at an angle of 450.

Now we calculate the shape of the trajectory in projectile motion. For this purpose we will express vertical displacement y as a function of horizontal displacement x, that is we will calculate a function of the type,

y = f(x)

as most of us are familiar with this type of functions describing the type of a curve. For example:

y = a + b x           - straight line

y = a + b x +c x2           - parabola

We write Eqs. 2.3 and 2.4 as a set of equations

x = v0x t                  (M2.13)

y = v0y t – (g t2) / 2     (M2.14)

and substitute Eq.M2.13 into M2.14. After doing this we obtain equation

(M2.15)

and we write it in the form

y= b x + c x2    (M2.15a)

with

b = sinα / cosα   (M2.15b)

and

c = - g / (2v02 cos2α)  (M2.15c)

Equation M2.15a is the equation defining a parabola, therefore the trajectory of the projectile motion is a PARABOLA.

We can ask the question – for which value of horizontal displacement x, the vertical displacement y achieves its maximum? It can be answered with the standard procedure of finding the maximum of a function. From Eq. M2.15a we have

(M2.16)

We must have

b – c x = 0

so

x = b / c

After substituting into this equation for b and c Eqs. M2.15b and M2.15c we get

(M2.17)

Notice, that it is exactly (1/2) of the largest horizontal displacement given by Eq. M.10. This is another evidence of symmetry of the projectile trajectory. The maximum height is at the half of the maximum displacement, for any angle of firing α

Often, as an example of a projectile an artillery shell is given. In this case, the initial velocity is too large to neglect the influence of air and therefore it is not a good example. Later on you will learn that resistance of air is approximately proportional to the square of the speed of a moving object. That is the reason why the motion of a bullet shot  from a rifle or any other gun cannot be described by equations governing the projectile motion.