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This carries on our exploration of projectile motion – this time we will explore what happens if gravity is not fixed, but is instead a function of time.  (This idea was suggested by and worked through by fellow IB teachers Daniel Hwang and Ferenc Beleznay).   In our universe we have a gravitational constant – i.e gravity is not dependent on time.  If gravity changed with respect to time then the gravitational force exerted by the Sun on Earth would lessen (or increase) over time with all other factors remaining the same.

Interestingly time-dependent gravity was first explored by Dirac and some physicists have tried to incorporate time dependent gravity into cosmological models.  As yet we have no proof that gravity is not constant, but let’s imagine a university where it is dependent on time.

Projectile motion when gravity is time dependent

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We can start off with the standard parametric equations for projectile motion. Here v is the initial velocity, theta is the angle of launch, t can be a time parameter and g is the gravitational constant (9.81 on Earth).  We can see that the value for the vertical acceleration is the negative of the gravitational constant.  So the question to explore is, what if the gravitational constant was time dependent?  Another way to think about this is that gravity varies with respect to time.

Linear relationship

If we have the simplest time dependent relationship we can say that:

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where a is a constant.  If a is greater than 0 then gravity linearly increases as time increases, if a is less than 0 than gravity linearly decreases as time increases.  For matters of slight convenience I’ll define gravity (or the vertical acceleration) as -3at.  The following can then be arrived at by integration:

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This will produce the following graph when we fix v = 10, a = 2 and vary theta:

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Now we can use the same method as in our Projectile Motion Investigation II to explore whether these maximum points lie in a curve.  (You might wish to read that post first for a step by step approach to the method).

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therefore we can substitute back into our original parametric equations for x and y to get:

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We can plot this with theta as a parameter.  If we fix v = 4 and a =2 we get the following graph:

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Compare this to the graph from Projectile Motion Investigation II, where we did this with gravity constant (and with v fixed as 10):

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The Projectile Motion Investigation II formed a perfect ellipse, but this time it’s more of a kind of egg shaped elliptical curve – with a flat base.  But it’s interesting to see that even with time dependent gravity we still have a similar relationship to before!

Inverse relationship

Let’s also look at what would happen if gravity was inversely related to time.  (This is what has been explored by some physicists).

In this case we get the following results when we launch projectiles (Notice here we had to use the integration by parts trick to integrate ln(t)).  As the velocity function doesn’t exist when t = 0, we can define v and theta in this case as the velocity and theta value when t = 1.

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Now we use the same trick as earlier to find when the gradient is 0:

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Substituting this back into the parametric equations gives:

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The ratio v/a will therefore have the greatest effect on the maximum points.

v/a ratio negative and close to zero:

v = 40, a = -2000, v/a = -0.02

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This gives us close to a circle, radius v, centred at (0,a).

v = 1, a = -10, v/a = -0.1

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Here we can also see that the boundary condition for the maximum horizontal distance thrown is given by x = v(e).

v/a ratio negative and large:

v = 40, a = -2, v/a = -20.

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We can see that we get an egg shape back – but this time with a flatter bulge at the top and the point at the bottom.  Also notice how quickly the scale of the shape has increased.

v/a ratio n/a (i.e a = 0)

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Here there is no gravitational force, and so projectiles travel in linear motion – with no maximum.

Envelope of projectiles for the inverse relationship

This is just included for completeness, don’t worry if you don’t follow the maths behind this bit!

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Therefore when we plot the parametric equations for x and y in terms of theta we get the envelope of projectile motion when we are in a universe where gravity varies inversely to time.  The following graph is generated when we take v = 300 and a = -10.  The red line is the envelope of projectiles.

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A generalized power relationship

Lastly, let’s look at what happens when we have a general power relationship i.e gravity is related to (a)tn.  Again for matters of slight convenience I’ll look at the similar relationship -0.5(n+1)(n+2)atn.

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This gives (following the same method as above:

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As we vary n we will find the plot of the maximum points.  Let’s take the velocity as 4 and a as 2.  Then we get the following:

When n = 0:

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When n = 1:

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When n =2:

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When n = 10:

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We can see the general elliptical shape remains at the top, but we have a flattening at the bottom of the curve.

When n approaches infinity:

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We get this beautiful result when we let n tend towards infinity – now we will have all the maximum points bounded on a circle (with the radius the same as the value chosen as the initial velocity.  In the graph above we have a radius of 4 as the initial velocity is 4. Notice too we have projectiles traveling in straight lines – and then seemingly “bouncing” off the boundary!

If we want to understand this, there is only going to be a very short window (t less than 1) when the particle can upwards – when t is between 0 and 1 the effect of gravity is effectively 0 and so the particle would travel in a straight line (i.e if the initial velocity is 5 m/s it will travel 5 meters. Then as soon as t = 1, the gravity becomes crushingly heavy and the particle falls effectively vertically down.

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Projectile Motion III: Varying gravity

We can also do some interesting things with projectile motion if we vary the gravitational pull when we look at projectile motion.  The following graphs are all plotted in parametric form.

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Here t is the parameter, v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which on Earth we will take as 9.81 m/s2.


Say we take a projectile and launch it with a velocity of 10 m/s.  When we vary the angle of launch we get the folowing graphs:

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On the y axis we have the vertical height, and on the x axis the horizontal distance.  Therefore we can see that the maximum height that we achieve is around 5m and the maximum horizontal distance is around 10m.

Other planets and universal objects

We have the following values for the gravitational pull of various objects:

Enceladus (Moon of Saturn): 0.111 m/s2, The Moon: 1.62 m/s2,  Jupiter: 24.8 m/s2, The Sun: 274 m/s2, White dwarf black hole surface gravity: 7×1012m/s2.

So for each one let’s see what would happen if we launched a projectile with a velocity of 10 m/s.  Note that the mass of the projectile is not relevant (though it would require more force to achieve the required velocity).


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The Moon:

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The Sun:

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Black hole surface gravity:

This causes some issues graphically!  I’ll use the equations derived in the last post to find the coordinates of the maximum point for a given launch angle theta:

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Here we have v = 10 and g = 7×1012m/s2.  For example if we take our launch angle (theta) as 45 degrees this will give the coordinates of the maximum point at:

(7.14×10-12, 3.57×10-12).


We can see how dramatically life would be on each surface!  Whilst on Earth you may be able to throw to a height of around 5m with a launch velocity of 10 m/s., Enceladus  would see you achieve an incredible 450m.  If you were on the surface of the Sun then probably the least of your worries would be how hight to throw an object, nevertheless you’d be struggling to throw it 20cm high.  And as for the gravity at the surface of a black hole you wouldn’t get anywhere close to throwing it a nanometer high (1 billionth of a meter).


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Projectile Motion Investigation II

Another example for investigating projectile motion has been provided by fellow IB teacher Ferenc Beleznay.  Here we fix the velocity and then vary the angle, then to plot the maximum points of the parabolas.  He has created a Geogebra app to show this (shown above).  The locus of these maximum points then form an ellipse.

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We can see that the maximum points of the projectiles all go through the dotted elliptical line.  So let’s see if we can derive this equation.

Let’s start with the equations for projectile motion, usually given in parametric form:

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Here v is the initial velocity which we will keep constant, theta is the angle of launch which we will vary, and g is the gravitational constant which we will take as 9.81.

We can plot these curves parametrically, and for each given value of theta (the angle of launch) we will create a projectile motion graph.  If we plot lots of these graphs for different thetas together we get something like this:

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We now want to see if the maximum points are in any sort of pattern.  In order to find the maximum point we want to find when the gradient of dy/dx is 0.  It’s going to be easier to keep things in parametric form, and use partial differentiation.  We have:

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Therefore we find the partial differentiation of both x and y with respect to t.  (This simply means we can pretend theta is a constant).

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We can then say that:

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We then find when this has a gradient of 0:

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We can then substitute this value of t back into the original parametric equations for x:

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and also for y:

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We now have the parametric equations in terms of theta for the locus of points of the maximum points.  For example, with g= 9.81 and v =1 we have the following parametric equations:

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This generates an ellipse (dotted line), which shows the maximum points generated by the parametric equations below (as we vary theta):

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And here is the graph:

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We can vary the velocity to create a new ellipse.  For example the ellipse generated when v =4 creates the following graph:

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So, there we go, we have shown that different ellipses will be created by different velocities.  If you feel like a challenge, see if you can algebraically manipulate the parametric equations for the ellipse into the Cartesian form!

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.

projectile motion

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 Vi and the angle of elevation is θ, then we can split this into vector components in the x and y direction:

Vxi = Vicosθ (Vxi is the horizontal component of the initial velocity Vi)
Vyi = Visinθ  (Vyi is the vertical component of the initial velocity Vi)

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/s2 ) which gives gravitational acceleration.  Therefore we can create 2 equations giving the changing velocity in both the x direction (Vx) and y direction (Vy):

Vx = Vicosθ
Vy = Visinθ – 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:

Sx = x =  (Vicosθ)t
Sy = y =  (Visinθ)t – 0.5gt2

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:

Vy = Visinθ – 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 Sx = (Vicosθ)t.

Sx =  (Vicosθ)t
Sx =  (100cos60).17.7
Sx  ≈ 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.

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IB Revision

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If you’re already thinking about your coursework then it’s probably also time to start planning some revision, either for the end of Year 12 school exams or Year 13 final exams. There’s a really great website that I would strongly recommend students use – you choose your subject (HL/SL/Studies if your exam is in 2020 or Applications/Analysis if your exam is in 2021), and then have the following resources:

Screen Shot 2018-03-19 at 4.42.05 PM.pngThe Questionbank takes you to a breakdown of each main subject area (e.g. Algebra, Calculus etc) and each area then has a number of graded questions. What I like about this is that you are given a difficulty rating, as well as a mark scheme and also a worked video tutorial.  Really useful!

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The Practice Exams section takes you to ready made exams on each topic – again with worked solutions.  This also has some harder exams for those students aiming for 6s and 7s and the Past IB Exams section takes you to full video worked solutions to every question on every past paper – and you can also get a prediction exam for the upcoming year.

I would really recommend everyone making use of this – there is a mixture of a lot of free content as well as premium content so have a look and see what you think.

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