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?
(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:
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.
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):
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.
(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
h/V0 = 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!
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Vot = h
Vo/h = t
Think it should be t = h/Vo, but thanks for the nice simple proof.
thanks – it’s been updated!
Aim a bit low or you will miss.
Isn’t there something missing? I think you’re just showing that there is a time when the bullet and the monkey are the same height. That’s not so surprising. But at that exact time, the bullet must also travel the exact horizontal distance to the monkey. This is not trivial and depends on the angle of the shot, which does not appear in your calculation at all.
you have 2 equations for the y coordinate of the monkey and the bullet respectively in terms of t. Therefore when they are equal then we have a coordinate (t,y) which tells us the time when they are indeed at the same height. The angle can change – this will change the equation for the motion of the bullet and will give a different (t,y) solution.
I think there is a misunderstanding. I try to be more precise. You want to show that there is a collision. For a collision the same hight is not enough, they must be at the exact same place. Technically: (y_B, t)=(y_M, t) is not enough, you need to show that there is an t with (x_B, y_B, t)=(x_M, y_M, t). So you additionally need to proof that for your t=h/v_y also x_B(t)=x_M(t) holds (where x_M(t) is constant).
Ok thanks – I see what you mean now! I suppose we can say informally from a graph sketch of the y,x plane that as we have a quadratic (from the projectile motion of the bullet) and a vertical line (motion of the monkey), then if they intersect they will intersect only once, so if there is a time when they intersect this must be that intersection point. I agree it would be nicer to be more formal. It think it needs involving ideas on horizontal motion – I’ll have a think on this. Thanks for your comments!
I thought it through and I think here is the missing part:
At t=0 the shooter aims at the monkey. Therefore, for the starting velocity v=(v_x,v_y) we have: v_x/v_y = d/h, with d as horizontal distance between shooter and monkey. With your solution t=h/v_y is then x_B(t) = v_x * t = v_x * h/v_y = h * v_x/v_y = h * d/h = d.
Thanks!