American shot putter Michelle Carter won the gold medal in her discipline at the Rio Olympics on Friday, August 12. Her winning throw measured 20.63 meters (67 ft 8 in). This made me wonder: how fast does she throw the shot?
Luckily, this is a simple calculus problem, best visualized using vectors. Throwing an object is a special case of simple projectile motion. The distance an object travels down range from where it is launched depends on a few parameters: the initial speed, the angle of launch, the height from which it is launched. We want to figure out the initial speed, so it might seem that we can't get started, but since we know how far Carter threw the shot, we can go backwards to figure out the initial speed.
To get a sense of how this works here's the video, in case you missed it.
As you can see, Carter launches the shot at about a 45° angle. If you think about it, this is clearly optimal--launch it at a steeper angle or a shallower angle and the shot will land shorter. Carter is 5' 9" tall, in meters that's 1.75. When she releases the shot, it is above her head. Exactly how high isn't clear, but let's just make an estimate and say it's a half-meter above her head. So if we think of her feet as located at the origin of a coordinate system, the shot has initial position (0,2.25). The horizontal component of the shot's velocity is v · cos(45°) and the vertical component is v · sin(45°), where v is the initial speed we are trying to figure out. The trick to solving this problem is then to think of this in terms of classical physics: ignoring air friction (which we may assume is negligible), the only force acting on the shot once it is released is gravity, pulling it straight down. If we denote the position of the shot at time t by s(t) = (x(t), y(t)), then we have the differential equations x″(t) = 0, x′(0) = v/√2, x(0) = 0; y″(t) = - g, y′(0) = v/√2, y(0) = 2.25, where g =9.8 m/s².
These are very simple equations to solve. We discover, by integrating twice for each of x and y and using the initial conditions that x(t) = (v/√2)·t and y(t) = 2.25 + (v/√2)·t - (1/2)gt². How do we solve for v? We know how far the winning throw traveled: 20.63 m. That means that the time the shot was in the air was t = 20.63√2/v. We can then put this into the y equation: when the shot lands y = 0, so substituting our t value into y(t) and setting it equal to 0, we find v = 13.5 m/s. I've omitted the algebra here, but it's pretty simple arithmetic. By the way, that means the shot was in the air for 2.16 s, which you can confirm is correct by watching the video.
How fast is that in terms most Americans understand? It's an easy conversion to get that v = 30.2 mph. You may find this number surprising, assuming the speed would be faster, but consider this: the women's Olympic shot weighs 4 kg (8.8 lb). Imagine the strength it takes to accelerate an 8.8 lb ball to 30 mph. We are always impressed by a major league pitcher's ability to throw at 95 mph, but a baseball has a mass of only 0.145 kg. In that context, Carter's power becomes even more impressive.
Congratulations to Michelle Carter and the rest of the U.S. Olympic team.
UPDATE: (1:50pm, Monday, August 15) A reader commented on Facebook that it's not obvious that the optimal launch angle is 45°. Indeed, it's not obvious, and, truthfully, it's not correct. Launching a projectile from a cliff, for example, requires decreasing the launch angle to increase the range. That's a bit counterintuitive, but it's not hard to see that the optimal launch angle decreases as the initial height y(0) increases. In the case of a shot putter (or football quarterback, for example), the relative launch height compared to the total distance the shot travels is pretty small (about 10% in this case) and so the optimal launch angle is close to 45°. But, just to be careful, I reworked the calculations with this in mind to find the optimal launch angle is approximately 42°. Since all of this was an estimate anyway, and since the sine and cosine of 42° is not far from 1/√2, the 30 mph calculation stands as a good approximation.
