Projectile Motion

A projectile can be defined as any object that is thrown/projected into the air and is moving freely. This means it doesn’t have a power source driving it. A rock thrown in the air can be considered a projectile because once it leaves your hand, it is moving freely in the air. A remote-controlled plane cannot be considered a projectile because even though it is moving through the air, it has a power source (electric motor) driving it. On the other hand, a remote-controlled plane, which has a dead battery, thrown into the air can be considered a projectile because it doesn’t have a power source driving it.

Once a projectile is released (e.g. rock leaves the hand, ball leaves a cricket bat), there is no horizontal/forward force acting on it. This makes logical sense because since a projectile is something with no power source driving it, then it can’t have any force pushing it in the horizontal/forward direction whilst it’s in motion. Due to this, a projectile has 0 horizontal acceleration. Since F=ma and F (force) is 0, then a (acceleration) has to be 0 too. The only time a projectile experiences a horizontal/forward force, is when it is being launched. It is the moment when a rock is in a person’s hand and they are moving the hand forward; in the direction they want to throw the rock in. It is the moment when a ball is in contact with a cricket bat, and the bat is being swung forwards.

Ignoring air resistance (which most of the questions do), the only force acting on a projectile after it is launched is its weight force, which is the force due to gravity, F_g. This force is always constant at 9.8 ms-2 and always acts directly downwards.

Projectiles Launched Horizontally

When solving a projectile motion question, you always have to separate the velocity and displacement into its vertical and horizontal components, then solve them separately. For example, if you are using the equation v^2=u^2+2as, you can either only use the final vertical velocity, initial vertical velocity and vertical displacement, or only the final horizontal velocity, initial horizontal velocity and horizontal displacement. In your workings, it is always a good idea to indicate which component you are using. This can be done by adding a subscript to the variables. For example, final vertical velocity can be represented as v_v (v indicating vertical) and final horizontal velocity can be represented as v_H (H indicating horizontal).

An example of a projectile launched horizontally is if you’re at a top of a cliff which has a perfectly horizontal surface at the top, and you roll a ball over the edge.

This is considered as a projectile launched horizontally because, as the name suggests, it is launched directly horizontal, meaning its initial vertical velocity is 0 because it isn’t moving up or down the vertical axis. Another key point which applies to both projectiles launched horizontally and projectiles launched obliquely is that the final horizontal velocity, v_H is the same as the initial horizontal velocity, u_H. This is because of the fact we established earlier; once launched, projectiles have 0 horizontal acceleration. If there is no acceleration, then the velocity would remain the same.

Uses of Each Equation

The 3 most common equations used whilst solving projectile motion questions are v = u+at , s = ut+ 1/2 at^2 , v^2 = u^2 + 2as.

A key thing to note about all 3 of the equations above is that it is only useful to look at the vertical components of the equations (eg. for equation 1; final vertical velocity = initial vertical velocity + vertical acceleration x time). It is pointless to look at the horizontal components for the above equations because as we know from the definition of a projectile, it doesn’t have any horizontal acceleration, but all 3 equations above have acceleration as a variable, hence it is no use to look at the horizontal component.

Looking at only the vertical components for the first equation, we can find out the final vertical velocity of an object if we know its initial vertical velocity and the time it took to reach the ground, or we can identify any other variable as long as we know the other 2. In the case of projectiles launched horizontally, we know that the initial vertical velocity is 0, so the equation simplifies down to only v = at.

Looking at the vertical components of the second equation, we can utilise this equation to find out its vertical displacement, given its initial vertical velocity and time. In the case of projectiles launched horizontally, initial vertical velocity is 0, so equation simplifies to s = 1/2 at^2. If you know the time, you can find it's vertical displacement, and vice versa.

Looking at the vertical components of the third equation, we can find out the final vertical velocity when we only know the initial vertical velocity and its vertical displacement. Again, a projectile launched horizontally doesn’t have an initial vertical velocity, so u is 0, simplifying the equation down to v^2=2as. If you know the vertical displacement, you can find the final vertical velocity, and vice versa.

Effects of Air Resistance

Air resistance acts in the opposite direction of motion.

The size of the force of air resistance depends on a number of factors. These include:

• Speed of the projectile. The faster the projectile is travelling, the greater the drag (force of air resistance).

• Cross-sectional area of the projectile. Greater the area equates to a larger drag.

• Aerodynamic shape of the object. The more streamlined the projectile is, the less the drag.

• Density of the air. More dense the air is, the higher the drag.

The subtopic of air resistance is theory only. It doesn’t have any calculation because that is outside the scope of Year 12 Physics and the questions in textbooks/tests/exams assume no air resistance.