**War Maths – Projectile Motion**

Despite maths having a reputation for being a somewhat bookish subject, it is also an integral part of the seamier side of human nature and has been used by armies to give their side an advantage in wars throughout the ages. Military officers all need to have a firm grasp of kinematics and projectile motion – so let’s look at some War Maths.

Cannons have been around since the 1200s – and these superseded other siege weapon projectiles such as catapults which fired large rocks and burning tar into walled cities. Mankind has been finding ever more ingenious ways of firing projectiles for the best part of two thousand years.

The motion of a cannon ball can be modeled as long as we know the initial velocity and angle of elevation. If the initial velocity is V_{i} and the angle of elevation is θ, then we can split this into vector components in the x and y direction:

V_{xi} = V_{i}cosθ (V_{xi} is the horizontal component of the initial velocity V_{i})

V_{yi} = V_{i}sinθ (V_{yi} is the vertical component of the initial velocity V_{i})

Next we know that gravity will affect the motion of the cannonball in the y direction only – and that gravity can be incorporated using g (around 9.8 m/s^{2} ) which gives gravitational acceleration. Therefore we can create 2 equations giving the changing velocity in both the x direction (V_{x}) and y direction (V_{y}):

V_{x} = V_{i}cosθ

V_{y} = V_{i}sinθ – gt

To now find the distance traveled we use our knowledge from kinematics – ie. that when we integrate velocity we get distance. Therefore we integrate both equations with respect to time:

S_{x} = x = (V_{i}cosθ)t

S_{y} = y = (V_{i}sinθ)t – 0.5gt^{2}

We now have all the information needed to calculate cannon ball projectile questions. For example if a cannon aims at an angle of 60 degrees with an initial velocity of 100 m/s, how far will the cannon ball travel?

**Step (1)** We find out when the cannon ball reaches maximum height:

V_{y} = V_{i}sinθ – gt = 0

100sin60 – 9.8(t) = 0

t ≈ 8.83 seconds

**Step (2)** We now use the fact that a parabola is symmetric around the maximum – so that after 2(8.83) ≈ 17.7 seconds it will hit the ground. Therefore substitute 17.7 seconds into the equation for S_{x} = (V_{i}cosθ)t.

S_{x} = (V_{i}cosθ)t

S_{x} = (100cos60).17.7

S_{x} ≈ 885 metres

So the range of the cannon ball is just under 1km. You can use this JAVA app to model the motion of cannon balls under different initial conditions and also factor in air resistance.

There are lots of other uses of projectile motion – the game Angry Birds is based on the same projectile principles as shooting a cannon, as is stunt racing – such as Evel Knieval’s legendary motorbike jumps:

If you enjoyed this post you might also like:

Bridge Building Lesson Plan. A lesson to introduce a practical example of maths and engineering.

Langton’s Ant – Order out of Chaos How computer simulations can be used to model life.

**IB Maths Revision**

I’d strongly recommend starting your revision of topics from Y12 – certainly if you want to target a top grade in Y13. My favourite revision site is Revision Village – which has a huge amount of great resources – questions graded by level, full video solutions, practice tests, and even exam predictions. Standard Level students and Higher Level students have their own revision areas. Have a look!

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