MOTION OF PROJECTILES

Learning Outcomes
Contents for this page	Related topics
Introduction Frame of reference The trajectory Vertical motion Momentum of projectiles Additional questions	Conservation of momentum in two dimensions Energy Work Power	Data Glossary
After studying this section, you will (a) know what is meant by a projectile, (b) be able to describe a projectile's trajectory in words and in mathematical terms, and (c), know how momentum is conserved along the trajectory.
Helpful background knowledge
Introduction to vectors	Addition of vectors	Equations of motion

Introduction:

When a bullet is fired from a gun, it is subject to an accelerating force (due to the expansion of the gases formed by the explosion of the gunpowder) while it is confined to the barrel of the gun (). Once the bullet leaves the barrel, it is subjected to external forces that shape the path along which that bullet will move. The forces are:

The force of gravity, mg, where m is the mass of the bullet and g the acceleration due to gravity.
Complex frictional forces due to air resistance (known as DRAG) that depend on the velocity of the bullet, its shape, its rotation, the density of the air, and the local speed of sound.
Forces introduced by the earth's rotation (of interest for those firing long-range guns).

With the above in mind, we consider a projectile to be any object that is free to move in two dimensions under the influence of gravity. Examples of this are bullets and shells fired from firearms and cannons, cricket balls that are hit or thrown, sky divers, the list is endless!

For the purpose of this course, we will only consider projectiles moving in such a manner as not to be affected by air resistance or the earth's rotation. It simplifies the topic enormously!

Some terms that are frequently used in connection with projectiles are:

MUZZLE VELOCITY: (only applicable for projectiles fired from rifles, cannons and so on). This is the velocity at which the projectile leaves the gun barrel. We are more used to call this velocity the INITIAL VELOCITY, u, of the moving object, for example the velocity of a golf ball immediately after being struck by the club.
ANGLE OF PROJECTION: this is the angle of the initial direction that the projectile moves along. This angle is measured from the horizontal plane, and is also known as the ELEVATION.
TRAJECTORY: the path taken by the projectile during its motion.
RANGE: the horizontal distance between the point of projection (for example, the muzzle of a gun) and the point where the projectile meets any plane through the point of projection (this point is called the POINT OF IMPACT). Note that the point of impact is not necessarily on the horizontal plane, as the projectile may hit the ground at a point higher or lower than the point of projection.
TIME OF FLIGHT: The time taken, from the moment of the projectile starts its motion to the time it reaches the end of its trajectory.
MUZZLE ENERGY: (only applicable for projectiles fired from rifles, cannons and so on). The kinetic energy of the projectile on leaving the gun barrel.

Frame of reference

It is convenient to ignore that a point on the surface of the earth has an eastward velocity (464 m.s^-1 at the equator) due to the earth's rotation, and we therefore use a two-dimensional frame of reference, with the horizontal displacement plotted along the x-axis, and the vertical displacement plotted along the y-axis. The point of departure of the projectile is normally taken as the origin. The set of (x,y) points for various time intervals constitutes the trajectory, shown in red in the diagram on the right.

In this reference frame, upward motion is taken as positive, displacements are positive if they are directed upwards, and the acceleration due to gravity, g is negative.

The trajectory:

The downward acceleration due to the earth's gravity causes the trajectory to be curved. We will show below that in the absence of drag, the trajectory has the shape of a PARABOLA ().

The velocity, v, of the projectile at any time is a vector which can be resolved into two rectangular components. The horizontal component, v_x, is not affected by the earth's gravity () and, in the absence of drag, will remain constant. The vertical component however, v_y will be influenced by the gravitational acceleration, -g, and will change in both magnitude and direction during the flight of the projectile. If the angle of projection is α₀, and the initial velocity is u, then the components of that initial velocity are

Note that at the top of the trajectory, that is, at the point H, the vertical component of the velocity is zero, and the velocity of the projectile will equal the horizontal velocity component.

(You might want to refresh your memory on the four equations of motion.) After any time interval from the start, Δt, the horizontal distance travelled will be

s_x = u cosα₀Δt

And in the same time, the vertical distance travelled will be

s_y = u sinα₀Δt - gΔt²

A plot of s_y against s_x (or a plot of s_y against t, since s_x µ t) is a parabola, which is the shape of the trajectory.

The figure above displays the special case where the end of the trajectory, R is in the same horizontal plane as the point of departure, P of the projectile. The above equations enable us to calculate various features of the trajectory. Referring to the diagram above,

The greatest height reached:

The greatest height reached, H, occurs when the vertical component of the velocity, v_y = 0. Then

v_y² = 0 = u² sin²α₀ - 2gH

From which we get

The time required to reach the greatest height:

The time, T, taken to reach the top of the trajectory occurs when the vertical component of the velocity, u sinα₀ = 0. Then

0 = u sinα₀- gT, hence T = u sinα₀ / g.

The time of flight:

The time of flight is the time taken, T_f, for the projectile to achieve zero vertical displacement, in other words, when s_y = 0 = u sinα₀ T_f - ½gT_f².

From which we get T_f = 2 u sinα₀ / g.

Note that this is twice the time taken to reach the maximum height.

The range:

The range, R is the distance covered horizontally during the time of flight T_f. Thus

R = u cosα₀ T_f = 2 u² sinα₀ cosα₀ / g () = u² sin2α₀ / g.

The longest range will occur when sin2α₀ = 1, that is, when α₀ is 45º.

Vertical motion:

Velocity is a vector that can ve resolved into horizontal and vertical components. If an object is projected in a vertical direction, that is the angle of projection α₀ is 90°, sinα₀ = 1, cosα₀ = 0. Thus, the horizontal velocity component is zero, and we can easily apply the equations of motion, using the vertical component only. In the examinations, you can expect all sorts of variants of the examples given below.

Graphs of objects moving under gravity:

You will be expected to recognise or reproduce sketch graphs of bodies that are either projected upwards or that are falling from some height. You should therefore familiarise yourselves with these. Look at the worked example below.

Momentum and energy of projectiles:

The momentum, p, of a moving particle is the product of the mass, m, a scalar quantity, and the velocity,v, a vector quantity. Hence the momentum of a projectile p = mv will be a vector quantity, having both magnitude and direction. At any one time, t, the horizontal component of the velocity v_x and the vertical component v_y will be

The corresponding components of the momentum will be

We note that since the horizontal component of the velocity, is constant, that is, it does not vary with time, the horizontal component of the momentum is conserved along the trajectory. The vertical component will decrease in a linear fashion with time, reach a zero value at the top of the trajectory, and gradually decreases further (the direction of the velocity changes at the top of the trajectory).

The magnitude of the momentum after any time interval Δt will be

The direction of the momentum after any time interval Δt will be at an angle q from the horizontal such that

(If you have any problems with this, see Addition of vectors: Method of components.)

Additional questions

Equations of motion:

Analytical expressions for the motion of bodies undergoing uniform acceleration can be derived. These expressions are used in problems involving displacement (s), initial velocity (u), final velocity (v), acceleration (a) and the time interval under consideration (Δt).

Equation 1: v = u + aΔt
Equation 2: s = ½(v + u)Δt
Equation 3: s = ut + ½aΔt²
Equation 4: v² = u² + 2as