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This is a classic puzzle which is discussed in some more detail by the excellent Wired article.  The puzzle is best represented by the picture below.  We have a hunter who whilst in the jungle stumbles across a monkey on a tree branch.  However he knows that the monkey, being clever, will drop from the branch as soon as he hears the shot being fired.  The question is therefore, at what angle should the hunter aim so that he still hits the monkey?

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(picture from the Wired article – originally from a UCLA physics textbook)

The surprising conclusion is that counter to what you would expect, you should actually still aim at the monkey on the branch – and in this way your bullet’s trajectory will still hit the monkey as it falls.  You can see a video of this experiment at the top of the page.

You can use tracking software (such as the free software tracker ) to show this working graphically.  Tracker provides a video demo with the falling monkey experiment:

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As you can see from the still frame, we have the gun in the bottom left corner, lined up with the origin, the red trace representing the bullet and the blue trace representing the falling monkey.

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We can then generate a graph to represent this data.  The red line is the height of the bullet with respect to time.  The faint blue line (with yellow dots) is the height of the monkey with respect to time.  We can see clearly that the red line can be modeled as a quadratic.  The blue line should in theory also be a quadratic (see below):

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but in our model, the blue line is so flat as to be better modeled as a linear approximation – which is shown in pink.  Now we can use regression technology to find the equation of both of these lines, to show not only that they do intersect, but also the time of that intersection.

We have the linear approximation as y = -18.5t + 14.5
and the quadratic approximation as y = -56t2+39t +0.1

So the 2 graphs will indeed intersect when -18.5t + 14.6 = -56t2+39t +0.1

which will be around 0.45 seconds after the gun is fired.

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(A more humane version, also from Wired – where we can throw the monkey a banana)

Newtonian Mathematics

The next question is can we prove this using some algebra?  Of course!  The key point is that the force of gravity will affect both the bullet and the falling monkey equally (it will not be affected by the different weights of the two – see the previous post here about throwing cannonballs from the Leaning Tower of Pisa).  So even thought the bullet deviates from the straight line path lined up in the gun sights, the distance the bullet deviates will be exactly the same distance that the monkey falls.  So they still collide.  Mathematically we have:

The vertical height of the bullet given by:

y = V0t – 0.5gt2

Where V0 is the initial vertical speed, t is the time, g is the gravitational force (9.8)

The vertical height of the monkey is given by:

y = h – 0.5gt2

where h is the initial vertical height of the monkey.

Therefore these will intersect when:

V0t – 0.5gt2 = h – 0.5gt2
V0t = h
V0/h = t

And for any given non-zero value of V0 we will have a t value – which represents the time of collision.

Well done – you have successfully shot the monkey!

If you like this you might also like:

Throwing cannonballs off the Leaning tower of Pisa – why weight doesn’t affect falling velocity

War Maths – how cannon operators used projectile motion to win wars

 

How to Design a Parachute

This post is also inspired by the excellent book by Robert Banks – Towing Icebergs. This book would make a great investment if you want some novel ideas for a maths investigation.

The challenge is to design a parachute with a big enough area to make sure that someone can land safely on the ground. How can we go about doing this? Let’s start (as in the last post) with some Newtonian maths.

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Newton’s Laws:

For an object falling through the air we have:

psgV – pagV – FD = psVa

ps = The density of the falling object
pa = The density of the air it’s falling in
FD = The drag force
g = The gravitational force
V = The volume of the falling object
a = The acceleration of the falling object

Time to simplify things

Things look a little complicated at the moment – luckily we can make our lives easier through a little simplification. pa will be many magnitudes smaller than than ps – as the density of air is much smaller than the density of objects like cannonballs. Therefore we ignore this part of the equation, giving an approximate equation:

psgV – FD ≈ psVa

We now rewrite things to make it easier to substitute values in later.

psV = m, where m = mass of an object (as density x volume = mass)
This gives:

mg – FD ≈ ma

and as mg = W (mass x gravitation force = weight) we can rewrite this as:

W – FD ≈ (w/g)a

 Now, the key information to know when looking at a parachute design is the terminal velocity that will be reached when the parachute is open – that means the maximum velocity that a parachutist will potentially be hitting the ground traveling.

Now, when a person is traveling at terminal velocity their acceleration is 0, so we can set a = 0 in the equation above to give:

W – FD = 0

Now we need an equation for FD (the drag force).
FD = 0.5paCDAU2

where
pa = density of the air
CD = the drag coefficient
A = area of parachute
U = velocity

So

when the parachutist is traveling at their terminal velocity with the parachute open we have:

W – FD = 0
W = 0.5paCDAU2

OK, nearly there. Next thing to consider is what is the maximum velocity we want someone to be traveling when they hit the ground.  This is advised to be around 5 m/s – similar to jumping from a 2 metre ladder.  Much more than this and you would risk breaking a bone (or worse!)

So we are finally ready to solve our equation. We want to find what value of A (the area of the parachute) will make us land safely.

We have:

pa = 0.6kg/m3 (approximate density of air at 3000m)
CD = 1.40 (a calculated drag coefficient for an open parachute)
U = velocity = 5m/s (this is the maximum velocity we want to want to avoid injury)
W = 100kg (we will have this as the combined weight of the parachutist and the parachute)

So,

W = 0.5paCDAU2
100 = 0.5(0.6)(1.40)A(5)2
A = 9.5m2

So if we had a circular parachute with radius 1.7m it should slow us down sufficiently for us to land safely.

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