How Much Superpower Does It Take to Bash a Bullet in Midair?

In a trailer for Wonder Woman 1984, our hero is under attack from some goons with guns. At one point (around 1:20 in the clip), she grabs one of the guns and pushes the slide, causing it to eject a round. As the bullet pops up into the air (in slow motion), she hits it off to the side—maybe shooting it into something?

I’m always looking for stuff to analyze, and this is a great physics example. So now I get to do what I do. Using physics, I want to figure out how much (super) power Wonder Woman would need in order to hit that bullet in midair.

Let’s start by looking at the upward motion of the bullet as it leaves the gun. If you’ve worked out movie scenes with me before (like the one where we analyzed Spider-Man’s leaps), you know the drill: I use my favorite video analysis tool, Tracker, to mark the location of an object in each frame of the video and combine that with the frame rate to get position and time data. But this particular shot is a little tougher. Here are the problems:

I need to know the size of something in the picture to fix a distance scale, and I don’t. The best I’ve got is the gun and Wonder Woman’s hand, so we’ll have to estimate here.
The camera zooms and pans during the shot, and we have to factor that out. I can use a stationary object like the wall behind Wonder Woman as a reference point, but it’s only an approximation, since the wall is farther away. Due to parallax, the change in apparent motion for this and the bullet will be slightly different.
It’s not in real time. Clearly it would take a bullet less than 3 seconds to get to its highest point. I don’t know the actual time between frames, and as we’ll see below, it’s not constant through the whole shot.

But there’s one thing I do know: This takes place on Earth, and when an object on Earth moves with only the gravitational force pulling on it (ignoring air resistance), it has a constant downward acceleration of 9.8 m/s2, represented by the symbol g. That means its motion obeys the following kinematic equation (where y is vertical position, v is velocity, and t is time):

Illustration: Rhett Allain

Since this is a quadratic equation, a plot of vertical position versus time should have the shape of a parabola. Notice that there are three variables here: position, time, and acceleration. If I know two of these, I can solve for the third. In this case, however, we only know the acceleration (g); we don’t have scales for distance or time.

So to nail down the distance, I estimated the width of Wonder Woman’s wrist at about 5 cm. Next, to circumvent the time problem, I created an arbitrary time unit, which I called fake seconds. Here then is a plot of vertical position versus time in fake seconds:

For the first part of this motion, the shape is parabolic, which means the bullet is indeed moving up with a constant downward acceleration. But check out that jump in the plot at around 2 (fake) seconds. That’s not right. Oh well. We can still do some fun stuff with this data. I’m just going to ask some questions and then go over the answers.

How long was the bullet in the air? What’s the real time scale?

Let’s assume my estimation of the distance scale was mostly legit. Mostly. That means I can find the vertical acceleration from the quadratic fit of the vertical bullet motion. This acceleration will be in units of meters per fake seconds squared instead of m/s². But now, if I set this fake-time acceleration equal to the real acceleration, I can solve for the relationship between fake and real seconds:

Illustration: Rhett Allain

That says a fake second is only about a twentieth of a real second. Using this conversion, I get a total bullet rise time of 0.169 seconds. That seems a lot more realistic. Oh, this assumes a constant time rate (which is probably not true, but oh well).

What kind of power would Wonder Woman need to hit this bullet?

Now to the real question. In this motion, Wonder Woman has to pull her arm back and then swing it forward to hit the bullet in a very short time. All I need to do is estimate the change in energy of her arm and divide by the time to move the arm forward. Yes, power is a measure of the rate that energy changes—often measured in units of watts.

Illustration: Rhett Allain

If I look just at the forward swing (not the pull back), Wonder Woman moves her arm over a distance of about 0.5 meters (I will label this as Δx) in a time of 0.037 seconds (assuming my unit conversion works). If the arm starts at rest, I can find both the average velocity and the final velocity.

Illustration: Rhett Allain

With a time of 0.037 seconds, this gives a final velocity of 26.8 meters per second (about 60 mph). The energy for this moving arm would be kinetic energy:

Illustration: Rhett Allain

Now I just need to estimate the mass of her arm. Yes, this is pretty tough—especially since different parts of the arm have different speeds. I’m going to go with a mass of 2 kilograms. I think that’s fair. Using this, I get a final kinetic energy of 721 joules. If you aren’t familiar with energy units, it takes about 10 joules to lift a textbook off the floor and put it on a table.

Now that I have the energy and the time, I can calculate Wonder Woman’s power. This particular move requires 19,000 watts. Oh, for comparison, your house runs at about 2,000 to 5,000 watts (your results may vary). What about a person? Most humans can produce over 1,000 watts for very short periods—but nothing close to 20,000 watts.

Maybe you’ll have a more intuitive feel for this if I convert that to horsepower. One hp is equal to about 746 watts. That means Wonder Woman would need about 25 hp to hit this bullet. That’s crazy high, but not super crazy. Especially for a superhero. But wait!

How fast is the bullet moving after she hits it?

I don’t know if Wonder Woman means to weaponize the bullet by hitting it. But let’s see how much speed she puts on it. The bullet is visible in one frame after impact, and It moves about 0.39 meters in a time of 0.0024 (real) seconds. That gives us a speed of 165 m/s (369 mph). Fast, but not really bullet-fast. To use more common bullet-speed units, that’s 541 feet per second. Most bullets go about twice that fast (or faster).

Maybe Wonder Woman wasn’t trying to shoot the bullet, but only knock it away. Guess we’ll have to wait till the movie comes out in June to find out!