Olympic Physics

This week the Olympics used the same arena that the original Olympians were at thousands of years ago. I saw them doing the Shot Put. Which got me thinking, of course, what is the best angle to lob a heavy rock in order to get the most distance.
If we start with some simple assumptions we can tackle this problem with algebra. It’s always good to know some boundary conditions just to make sure you can validate your answer. So lets start with these assumptions: We’re launching from ground level on a perfectly level field and there is no air friction. Oh, and as soon as the shot put hits the ground it stops. I think we can all agree that given those conditions, a perfectly horizontal launch won’t go anywhere because it never leaves the ground. Likewise a perfectly vertical launch net any distance either because it will land right where it started. Finally, let’s just take an educated guess: It is going to be somewhere between horizontal and vertical — let’s say 45 degrees as a first approximation.
Now let’s tackle the Algebraic solution to this puzzle. First, I always draw a picture to help me keep track of what’s going on.
The free body diagram.
Here I’ve shown the path of the ball as an arc. The total time (indicated by the stopwatch on the right) is t seconds. The length of the throw is X meters. The initial velocity of the ball is V meters per second at some angle, A, above horizontal. We’ll consider two parts of the velocity: The horizontal speed (Vx) and the initial vertical speed (Vy).
Here is the equation we need to solve this: d = vt + ½at2. We’ll also need to know the acceleration of gravity, G.
Well, the time the ball stays in the air is a function of the vertical component of the throw. So using the above equation we know v is actually Vy. The acceleration, a, is simply G. Finally, the ball is at ground level when the distance travelled is zero. Thus we want to solve:

0 = Vyt - ½Gt2

which has two solutions : t = 0 and t = 2 Vy / G . Hang on, we are making progress!
Using our original equation again, we can see how far the ball has traveled in that time:
X = Vxt

X = 2 VxVy / G

Now we simply use a little trig to relate Vx and Vy to V and A.
Vx = V cos(A)

Vy = V sin(A)

Thus our solution can be shown as this, in terms of V, A and G:
X = 2 V2 cos(A) sin(A) / G

The interesting thing about this result is that it shows that the distance a ball will travel is proportional to the square of how fast you throw it. This means that it is actually easier to differentiate how fast two atheletes can throw a ball based on how far the ball travels than if you were to measure the speed directly. Also, this makes sense in terms of energy, since we know that kinetic energy is proportional to the square of speed.
Moving on with the algebra… We can assume:
sin(2x) = 2 sin(x)cos(x)

That’s some trig I dug up on on the internet. Which gives us.
X = V2 sin(2A) / G

So what it all boils down to is finding the maxima of sin(2A). You need calculus — just a little tiny bit — to do this. What you do is look for a spot on the curve where it’s flat. It gets a little hairy, so I’ll just show you how to set it up and then solve it for you:
dX/dA = 0 = 2 V2 cos(2A)

Which basically boils down to the angle where cos(2A) = 0. The simplest answer is: 45°.
The problem really gets interesting when you factor in the height of the shot put upon throw (since the ball is several feet of the ground when the athelete releases it). Just how much this matters is left as an exercise for the reader. 🙂

6 replies on “Olympic Physics”

Whew, I didn’t realize I had to worry about so much while watching the Olympics. I am guessing most of us would drop it on our foot, despite your calculations. Now, do something with the physics of beach volleyball (never mind the uniforms in that formula) which NBC seems totally enthralled with.

Actually, the physics of beach volleyball I would be interested in is how those skimpy suits stay on while the athletes leap, twist, and dive into the sand. There’s a feat of engineering for you.
Also, Andy, can you calculate the impact of the “AAAEEEEAAAUGH!” scream in shotputting? It must make a difference, since all the shotputters seem to do it!

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